Unit 1: Functions

Theme A

Foundations of functions

Before anything else in precalculus, you need a precise answer to two questions: what is a function, and how do we write down what it does. Everything in the rest of the unit is built on these.

Objective 01 / 16

Use multiple representations to identify functions and justify what defines a function

A function from a set of inputs (the domain) to a set of outputs is a rule that assigns to each input exactly one output. That is the entire definition, and the phrase that does the work is "exactly one."

Two different inputs are allowed to share the same output.
One input is never allowed to produce two different outputs.

The five representations

You should be able to test the definition in every form a function can appear:

Ordered pairs / a set: $\{(1,2),(2,4),(3,2)\}$ — check that no first coordinate repeats with a different second coordinate.
Table: scan the input column — no input value may appear twice with different outputs.
Mapping diagram: exactly one arrow must leave each input.
Equation: solve for $y$. If a single $x$ can force $y=\pm(\dots)$, it is not a function of $x$.
Graph: the Vertical Line Test — every vertical line meets the graph at most once.

"Justify what defines a function" means you must say which rule is violated, not just answer yes or no. A complete justification names the input that has two outputs, or states that every input has exactly one.

Worked example

Decide whether each represents $y$ as a function of $x$.

$\{(-2,1),(0,1),(3,1)\}$ — Function. Each input $-2,0,3$ appears once. The output $1$ repeating is fine.
$\{(1,2),(1,3),(4,5)\}$ — Not a function. The input $1$ is paired with both $2$ and $3$.
$x^2 + y^2 = 25$ — Solve: $y = \pm\sqrt{25-x^2}$. At $x=0$, $y = \pm 5$. Not a function (its graph, a circle, fails the Vertical Line Test).
$y = |x|$ — every $x$ gives one $y$. Function.

Watch out

Repeated outputs are legal; repeated inputs are not. A horizontal-looking table (all outputs equal) is still a function.
A relation that "passes through $(2,5)$ and $(2,-5)$" is automatically not a function.
"It's a function because it has an equation" is not a justification. $x = y^2$ has an equation and is not a function of $x$.

Practice

Is $\{(4,1),(5,2),(4,3)\}$ a function? Justify.
Does $x = y^2 - 1$ define $y$ as a function of $x$? Justify with a specific input.
A mapping sends $a\to 1$, $b\to 2$, $c\to 2$. Function? What if we also add $a\to 3$?
The graph of a horizontal line $y = -2$: function or not, and why?
A table has inputs $1,2,3,2$ with outputs $5,7,9,7$. Function?

Answer key

Not a function — input $4$ maps to both $1$ and $3$.
No — at $x = 0$, $y^2 = 1$ so $y = \pm 1$; the input $0$ has two outputs.
Yes, it is a function (each of $a,b,c$ has one arrow). Adding $a\to 3$ breaks it — $a$ now has two outputs.
A function — every vertical line hits it exactly once; every input gives the single output $-2$.
Yes — the input $2$ appears twice but with the same output $7$, so no input has two different outputs.

Your turn: write an exam question for this objective — e.g. give a four-row table with one repeated input and ask "function or not? justify."

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 02 / 16

Apply function notation to evaluate a function at a value or expression

The symbol $f(x)$ is read "$f$ of $x$" and means the output of the function $f$ when the input is $x$. It is not $f$ multiplied by $x$. The letter in the parentheses is a placeholder: whatever you put in the parentheses on the left, you put in every slot on the right.

Evaluating at a value — substitute a number:

If $f(x) = x^2 - 3x + 1$, then $f(4) = (4)^2 - 3(4) + 1 = 16 - 12 + 1 = 5$, and $f(-2) = 4 + 6 + 1 = 11$.

Evaluating at an expression — substitute the whole expression, wrapped in parentheses, then simplify:

\[ f(a+1) = (a+1)^2 - 3(a+1) + 1 = a^2 + 2a + 1 - 3a - 3 + 1 = a^2 - a - 1 \] \[ f(2x) = (2x)^2 - 3(2x) + 1 = 4x^2 - 6x + 1 \]

The difference quotient $\dfrac{f(x+h) - f(x)}{h}$ is the single most important "evaluate at an expression" task in the course — it is the seed of the derivative.

Worked example — difference quotient

For $f(x) = x^2 - 3x + 1$, simplify $\dfrac{f(x+h) - f(x)}{h}$.

$f(x+h) = (x+h)^2 - 3(x+h) + 1 = x^2 + 2xh + h^2 - 3x - 3h + 1$
$f(x+h) - f(x) = (x^2 + 2xh + h^2 - 3x - 3h + 1) - (x^2 - 3x + 1) = 2xh + h^2 - 3h$
Divide by $h$: $\dfrac{h(2x + h - 3)}{h} = 2x + h - 3$

Watch out

$f(-x)$ means "substitute $-x$"; $-f(x)$ means "negate the output." They are usually different.
$f(x+h) \neq f(x) + h$, and $(x+h)^2 \neq x^2 + h^2$. Expand fully.
If $g(x) = \dfrac{5}{x}$, then $g(0)$ is undefined — write "undefined," not $0$.
Reading from a graph: $f(0)$ is the $y$-intercept; solving $f(x) = 0$ gives the $x$-intercepts. Don't swap them.

Practice

Let $f(x) = 2x^2 + x$ and $h(x) = \sqrt{x+4}$.

$f(3)$ and $f(-1)$
$f(t)$ and $f(t-1)$
The difference quotient of $f$, fully simplified
$h(5)$, $h(-4)$, and $h(a^2)$
For $f$: is $f(-2)$ equal to $-f(2)$?

Answer key

$f(3) = 18 + 3 = 21$; $f(-1) = 2 - 1 = 1$.
$f(t) = 2t^2 + t$; $f(t-1) = 2(t-1)^2 + (t-1) = 2t^2 - 4t + 2 + t - 1 = 2t^2 - 3t + 1$.
$f(x+h) = 2x^2 + 4xh + 2h^2 + x + h$; minus $f(x)$ gives $4xh + 2h^2 + h$; divide by $h$: $4x + 2h + 1$.
$h(5) = 3$; $h(-4) = 0$; $h(a^2) = \sqrt{a^2 + 4}$.
$f(-2) = 8 - 2 = 6$; $-f(2) = -(8 + 2) = -10$. Not equal.

Your turn: write a question that forces the parentheses habit — e.g. "given $f(x)=3x^2-x$, find $f(x-2)$ and the difference quotient."

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Theme B

Behavior, domain, and meaning

Now that a function is defined, we describe it: where it rises and falls, which inputs and outputs are allowed, and what its values mean when the function models something real.

Objective 03 / 16

Determine increasing and decreasing intervals of various functions

Read the graph left to right, like reading a sentence.

Increasing on an interval: as $x$ grows, $y$ grows. Formally, if $x_1 < x_2$ then $f(x_1) < f(x_2)$.
Decreasing: as $x$ grows, $y$ falls. If $x_1 < x_2$ then $f(x_1) > f(x_2)$.
Constant: $y$ does not change.

Report the answer as intervals of $x$, written with open intervals joined by $\cup$. The turning points themselves are left out — at a peak or valley the function is neither increasing nor decreasing. Where the behavior switches, you have a local maximum (increasing → decreasing) or local minimum (decreasing → increasing).

Worked examples

$f(x) = x^2 - 4x + 1$ has its vertex at $x = 2$. Decreasing on $(-\infty, 2)$, increasing on $(2, \infty)$.
$f(x) = |x - 1|$: decreasing on $(-\infty, 1)$, increasing on $(1, \infty)$.
A curve with a local maximum at $x = -1$ and a local minimum at $x = 3$: increasing on $(-\infty, -1) \cup (3, \infty)$, decreasing on $(-1, 3)$.
A piecewise graph that rises to $x = 0$, is flat until $x = 4$, then falls: increasing on $(-\infty, 0)$, constant on $(0, 4)$, decreasing on $(4, \infty)$.

Watch out

Answer with $x$-intervals, never $y$-values and never a single point.
Join intervals with $\cup$ (union), not $\cap$.
"Increasing" describes a direction of travel. A function can be increasing while its values are still negative.
A function is not "increasing at $x = 3$" — increasing/decreasing are properties of intervals.

Practice

$g(x) = (x + 3)^2 - 5$: state the increasing and decreasing intervals.
A cubic-shaped curve has a local max at $(-2, 6)$ and a local min at $(1, -2)$. Give all intervals.
$f(x) = -x^2 + 6x$: where is it increasing?
A graph is a line of slope $2$ for $x < 1$, horizontal for $1 \le x \le 5$, and a line of slope $-1$ for $x > 5$. Describe its behavior.

Answer key

Vertex at $x = -3$: decreasing on $(-\infty, -3)$, increasing on $(-3, \infty)$.
Increasing on $(-\infty, -2) \cup (1, \infty)$; decreasing on $(-2, 1)$.
Vertex at $x = 3$; opens down, so increasing on $(-\infty, 3)$.
Increasing on $(-\infty, 1)$, constant on $(1, 5)$, decreasing on $(5, \infty)$.

Your turn: sketch (or describe) a graph with two turning points and ask for all three interval types.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 04 / 16

Identify the domain and range of various functions using interval, set, and/or inequality notation

The domain is every input the function is allowed to take; the range is every output it actually produces.

Three notations for the same set

Interval: $(-\infty, 3\,]$, $(0, 5)$, $[2, \infty)$. A square bracket includes the endpoint; a parenthesis excludes it; $\infty$ always gets a parenthesis.
Set-builder: $\{x \mid x \le 3\}$, $\{x \mid x \neq 0\}$.
Inequality: $x \le 3$, or $0 < x < 5$.

Finding the domain from a formula — start with "all real numbers," then remove what breaks the math:

Division by zero: $f(x) = \dfrac{1}{x - 2}$ needs $x \neq 2$: $(-\infty, 2) \cup (2, \infty)$.
Even root of a negative: $f(x) = \sqrt{x - 5}$ needs $x - 5 \ge 0$: $[5, \infty)$.
Both at once: $f(x) = \dfrac{\sqrt{x + 1}}{x - 3}$ needs $x \ge -1$ and $x \neq 3$: $[-1, 3) \cup (3, \infty)$.

From a graph: sweep left-to-right for the domain (the $x$-shadow), bottom-to-top for the range (the $y$-shadow). Watch open vs. closed dots and arrows that mean "continues forever."

Worked examples

$f(x) = x^2$: domain $(-\infty, \infty)$; range $[0, \infty)$ — squares are never negative.
$f(x) = \sqrt{x}$: domain $[0, \infty)$; range $[0, \infty)$.
$f(x) = \dfrac{1}{x}$: domain $(-\infty, 0) \cup (0, \infty)$; range $(-\infty, 0) \cup (0, \infty)$.
Top half of the circle $x^2 + y^2 = 9$: domain $[-3, 3]$; range $[0, 3]$.

Watch out

$\infty$ is never bracketed: write $(2, \infty)$, never $(2, \infty\,]$.
Use $\cup$ to show a gap; a comma inside one interval means something else.
Range is usually the harder half — reason from the function's shape (minimum value, asymptote, end behavior), not just the formula.
An odd root, e.g. $\sqrt[3]{x}$, has domain all reals — only even roots restrict.

Practice

Give domain and range in interval notation.

$f(x) = \dfrac{2}{x + 4}$ (domain only)
$f(x) = \sqrt{7 - x}$ (domain only)
$f(x) = (x - 2)^2 + 3$
$f(x) = -|x| + 5$
$f(x) = \dfrac{x}{\sqrt{x - 1}}$ (domain only)

Answer key

$x \neq -4$: $(-\infty, -4) \cup (-4, \infty)$.
$7 - x \ge 0 \Rightarrow x \le 7$: $(-\infty, 7\,]$.
Domain $(-\infty, \infty)$; minimum value $3$ at $x = 2$, opens up, so range $[3, \infty)$.
Domain $(-\infty, \infty)$; maximum value $5$, opens down, so range $(-\infty, 5\,]$.
Need $x - 1 > 0$ (strict, because it's also in a denominator): $(1, \infty)$.

Your turn: build one function that combines a square root and a denominator, and ask for the domain in all three notations.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 05 / 16

Determine function values and inverse values and connect their meaning with a given scenario or model

When a function models a real situation, an equation is also a sentence. Suppose $C(t)$ gives the number of subscribers after $t$ months.

$C(5) = 800$ reads: "After 5 months there are 800 subscribers." Input = months, output = subscribers.
$C^{-1}(800) = 5$ reads: "It takes 5 months to reach 800 subscribers." Input = subscribers, output = months.

The inverse swaps the roles of input and output — and therefore swaps the units. Finding an "inverse value" $C^{-1}(k)$ means solving the equation $C(t) = k$ for $t$; you do not need a formula for $C^{-1}$.

Worked example

A tank drains so that $V(t) = 500 - 20t$ liters remain after $t$ minutes.

$V(10) = 500 - 200 = 300$: after 10 minutes, 300 L remain.
$V^{-1}(400)$: solve $500 - 20t = 400 \Rightarrow t = 5$. The tank holds 400 L after 5 minutes.
$V^{-1}(0)$: solve $500 - 20t = 0 \Rightarrow t = 25$. The tank is empty at 25 minutes.

Watch out

$f^{-1}(a)$ is not $\dfrac{1}{f(a)}$. It is the input that produces the output $a$.
State the answer as a full sentence with the correct nouns and units — "5 minutes," not just "5."
Read tables in the right direction: $f(2010)$ looks up the year in the input column; $f^{-1}(50000)$ looks up $50000$ in the output column.

Practice

$H(t) = 3 + 2t$ is the height in cm of a plant $t$ weeks after planting.

Compute $H(6)$ and write its meaning.
Find $H^{-1}(15)$ and write its meaning.
Interpret $H^{-1}(3) = 0$.
A population table shows $(2015, 4000)$ and $(2020, 6000)$. What do $P(2020)$ and $P^{-1}(4000)$ mean?

Answer key

$H(6) = 15$ cm — six weeks after planting the plant is 15 cm tall.
$3 + 2t = 15 \Rightarrow t = 6$ weeks — it takes 6 weeks to reach 15 cm.
At $t = 0$ (planting time) the height was 3 cm.
$P(2020) = 6000$: the population in 2020 was 6000. $P^{-1}(4000) = 2015$: the population was 4000 in the year 2015.

Your turn: give a linear model with units and ask for one function value and one inverse value, each explained in a sentence.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Theme C

Composition of functions

Composition chains two functions so the output of one becomes the input of the next. The new subtlety is the domain: an input must survive both steps.

Objective 06 / 16

Determine the composite function from a given pair of functions and their associated domain

$(f \circ g)(x) = f\big(g(x)\big)$: apply $g$ first, then feed the result into $f$. Read the symbol right-to-left. In general $(f \circ g) \neq (g \circ f)$, and neither equals the product $f \cdot g$.

Domain of a composite — two conditions

$x$ must be in the domain of the inner function $g$.
$g(x)$ must be in the domain of the outer function $f$.

The second condition can add restrictions that disappear when you simplify the formula — so find the domain before simplifying.

Worked example — order and domain

$f(x) = \sqrt{x}$, $g(x) = x - 3$.

$(f \circ g)(x) = \sqrt{x - 3}$. Need $x - 3 \ge 0$: domain $[3, \infty)$.
$(g \circ f)(x) = \sqrt{x} - 3$. Need $x \ge 0$: domain $[0, \infty)$. Different function, different domain.

Worked example — a hidden restriction

$f(x) = \dfrac{1}{x - 1}$, $g(x) = \dfrac{1}{x}$.

Inner domain: $x \neq 0$.
$(f \circ g)(x) = \dfrac{1}{\frac{1}{x} - 1} = \dfrac{1}{\frac{1 - x}{x}} = \dfrac{x}{1 - x}$.
Also need $g(x) \neq 1$, i.e. $\dfrac{1}{x} \neq 1$, so $x \neq 1$.
Domain: $x \neq 0$ and $x \neq 1$ → $(-\infty, 0) \cup (0, 1) \cup (1, \infty)$.

The simplified form $\dfrac{x}{1 - x}$ alone only shows $x \neq 1$; the $x \neq 0$ from the inner function must be carried along by hand.

Watch out

$(f \circ g)(x) = f(g(x))$, not $g(f(x))$, and not $f(x)\cdot g(x)$.
Wrap $g(x)$ in parentheses before substituting into $f$.
Restrictions from the inner function stay, even if the algebra "cleans up."

Practice

$f(x) = x^2 + 1$, $g(x) = 2x - 5$, $k(x) = \sqrt{x}$.

$(f \circ g)(x)$ and its domain
$(g \circ f)(x)$ and its domain
$(f \circ k)(x)$ and its domain
$(k \circ g)(x)$ and its domain

Answer key

$(2x - 5)^2 + 1 = 4x^2 - 20x + 26$; domain $(-\infty, \infty)$.
$2(x^2 + 1) - 5 = 2x^2 - 3$; domain $(-\infty, \infty)$.
$(\sqrt{x})^2 + 1 = x + 1$, but the inner $\sqrt{x}$ needs $x \ge 0$: domain $[0, \infty)$.
$\sqrt{2x - 5}$; need $2x - 5 \ge 0$: domain $\left[\tfrac{5}{2}, \infty\right)$.

Your turn: pick an inner function with a domain restriction that cancels in the simplified composite, and ask for the domain.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 07 / 16

Evaluate composite functions at a constant, taking the domain into account

To find $(f \circ g)(a)$: compute $g(a)$, then apply $f$ to that number. Always check that the intermediate value $g(a)$ is a legal input to $f$. If it isn't, the answer is undefined — and "undefined" is a correct, complete answer.

This works from formulas, from tables, or from graphs: for tables, look up $g(a)$ in $g$'s table, then look that value up in $f$'s table.

Worked example — from formulas

$f(x) = \sqrt{x}$, $g(x) = 5 - x$.

$(f \circ g)(1) = f(g(1)) = f(4) = 2$. ✓
$(f \circ g)(9) = f(g(9)) = f(-4) = \sqrt{-4}$ → undefined. So $9$ is not in the domain of $f \circ g$, even though the first step worked.
$(g \circ f)(16) = g(f(16)) = g(4) = 1$. ✓

Worked example — from tables

$g$: $g(0)=2,\; g(1)=5,\; g(2)=3$. $f$: $f(2)=7,\; f(3)=-1,\; f(5)=0$.

$(f \circ g)(0) = f(g(0)) = f(2) = 7$.
$(f \circ g)(2) = f(g(2)) = f(3) = -1$.
$(f \circ g)(1) = f(g(1)) = f(5) = 0$.
$(f \circ g)(2)$ would be undefined if $g(2)$ were, say, $4$ — because $f(4)$ is not defined by the table.

Watch out

Do the steps in the right order: inner first.
Check the middle number against the outer function's domain every time.
Don't invent an $f$-value that a table or graph doesn't give you.

Practice

$p(x) = x^2 - 9$, $q(x) = \sqrt{x}$.

$(q \circ p)(5)$
$(q \circ p)(1)$
$(p \circ q)(16)$
Using the tables above, $(g \circ f)(3)$ and $(f \circ f)(2)$.

Answer key

$p(5) = 16$, then $q(16) = 4$.
$p(1) = -8$, then $q(-8) = \sqrt{-8}$ → undefined.
$q(16) = 4$, then $p(4) = 7$.
$(g \circ f)(3) = g(f(3)) = g(-1)$ → undefined (no $g(-1)$ in the table). $(f \circ f)(2) = f(f(2)) = f(7)$ → undefined (no $f(7)$).

Your turn: write a table pair where exactly one of $(f\circ g)(a)$, $(g\circ f)(a)$ is undefined, and ask the student to find both.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Theme D

Properties and classification

Two independent ways to sort functions: by symmetry (even, odd, neither) and by how the mapping behaves (injective, surjective, bijective). The second one decides whether an inverse can exist.

Objective 08 / 16

Identify function properties (even, odd, neither) and types of functions (injective, surjective, bijective) from graphs, mappings, tables, equations, etc.

Symmetry: even, odd, or neither

Even: $f(-x) = f(x)$ for all $x$. Graph is symmetric across the $y$-axis. Examples: $x^2$, $|x|$, $x^4 - 3x^2$.
Odd: $f(-x) = -f(x)$ for all $x$. Graph has $180^\circ$ rotational symmetry about the origin. Examples: $x$, $x^3$, $\dfrac{1}{x}$.
Neither: fails both. Most functions are here — e.g. $f(x) = x^2 + x$.

The test: compute $f(-x)$, simplify, and compare it to $f(x)$ and to $-f(x)$.

Why "odd" means $180^\circ$ symmetry. Turning a graph $180^\circ$ about the origin sends every point $(x, y)$ to $(-x, -y)$. So the turned graph contains the point $(-x, -y)$ exactly when the original contains $(x, y)$ — i.e. exactly when $-y = f(-x)$, that is $f(-x) = -f(x)$. When that holds, the turn lands the graph back on itself:

rotation 0°

$y = x^3 - 3x$ turned $180^\circ$ about the origin. The hollow marker rides from $(a, f(a))$ around to $(-a, -f(a))$, and the whole curve lands back on the faint copy — that is what $f(-x) = -f(x)$ looks like. (An even function would instead land on itself after a mirror flip across the $y$-axis.)

Type: injective, surjective, bijective

These describe how the function maps its domain onto a stated codomain (the "target" set you're told to use).

Injective (one-to-one): different inputs always give different outputs. Graph test: the Horizontal Line Test — every horizontal line meets the graph at most once. A strictly increasing or strictly decreasing function is automatically injective.
Surjective (onto): every element of the codomain is actually hit — the range equals the codomain.
Bijective: both injective and surjective. Exactly the bijective functions have an inverse function (on that domain and codomain).

Worked example — symmetry

$f(x) = x^4 - 2x^2$: $f(-x) = x^4 - 2x^2 = f(x)$ → even.
$f(x) = x^3 - x$: $f(-x) = -x^3 + x = -(x^3 - x) = -f(x)$ → odd.
$f(x) = x^3 + 1$: $f(-x) = -x^3 + 1$; $-f(x) = -x^3 - 1$. Matches neither → neither.

Worked example — type depends on the codomain

$f:\mathbb{R}\to\mathbb{R},\; f(x) = x^3$: passes HLT (injective); range is all of $\mathbb{R}$ (surjective) → bijective.
$f:\mathbb{R}\to\mathbb{R},\; f(x) = x^2$: $f(2) = f(-2)$ (not injective); negatives are never outputs (not surjective) → neither.
$f:[0,\infty)\to[0,\infty),\; f(x) = x^2$: now injective and surjective → bijective.
$f:\mathbb{R}\to\mathbb{R},\; f(x) = e^{x}$: injective, but never $\le 0$, so not surjective onto $\mathbb{R}$.

Watch out

"Odd function" is not the same as "odd-degree polynomial" — $x^3 + 1$ has odd degree but is neither.
Surjectivity is meaningless without a stated codomain. If you're not told one, you can only discuss injectivity.
The Vertical Line Test checks "is it a function"; the Horizontal Line Test checks "is it one-to-one." Don't mix them up.
In a mapping diagram: injective = no target has two arrows in; surjective = no target has zero arrows in.

Practice

Classify by symmetry: (a) $f(x) = 5x^2 - 1$, (b) $f(x) = x^5 + x^3$, (c) $f(x) = x^2 + 4x$, (d) $f(x) = \dfrac{3}{x}$.
$f:\mathbb{R}\to\mathbb{R},\; f(x) = 2x - 7$: injective? surjective? bijective?
$f:\mathbb{R}\to\mathbb{R},\; f(x) = |x|$: which of the three types, if any?
A mapping sends $1\to a,\; 2\to b,\; 3\to a$ with codomain $\{a, b, c\}$. Injective? Surjective?

Answer key

(a) even; (b) odd; (c) neither ($f(-x) = x^2 - 4x$); (d) odd.
Strictly increasing → injective; range is all of $\mathbb{R}$ → surjective; therefore bijective.
$|{-3}| = |3|$ → not injective; negatives are never outputs → not surjective onto $\mathbb{R}$; neither.
$1$ and $3$ both map to $a$ → not injective. $c$ is never hit → not surjective.

Your turn: give the same rule $f(x)=x^2$ with two different codomains and ask how the classification changes.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Theme E

Inverse functions

An inverse runs a function backwards. Three questions, in order: can it be reversed, how do you reverse it, and how are the two related once you have both.

Objective 09 / 16

Predict when a function will have an inverse

A function has an inverse function if and only if it is one-to-one (injective) on its domain — no output value is used twice. Equivalent ways to check:

The graph passes the Horizontal Line Test.
The function is strictly increasing or strictly decreasing over its whole domain (a common sufficient condition in this course).
In a table or mapping, no output repeats.

If a function is not one-to-one, you can often restrict the domain to a piece where it is monotonic, and an inverse exists on that piece.

Worked example

$f(x) = x^3$: passes HLT → invertible on all of $\mathbb{R}$.
$f(x) = x^2$: fails HLT ($f(2) = f(-2)$) → no inverse. Restrict to $x \ge 0$ → invertible there.
$f(x) = 2^{x}$: strictly increasing → invertible (its inverse is $\log_2 x$).
$f(x) = |x|$: not one-to-one → restrict to $x \ge 0$ (or $x \le 0$).
$f(x) = 7$ (constant): every output is $7$ → definitely not invertible.

Watch out

Passing the Vertical Line Test only means "it's a function." Invertibility needs the Horizontal Line Test.
A parabola, $|x|$, $x^4$, $\cos x$, and similar "U-shaped or wavy" graphs are never one-to-one on their full natural domain.
When you restrict the domain, say which piece you kept.

Practice

State whether each has an inverse on the given domain; if not, give a restriction that works.

$f(x) = 4x - 1$, domain $\mathbb{R}$
$f(x) = (x - 3)^2$, domain $\mathbb{R}$
$f(x) = \sqrt{x}$, domain $[0, \infty)$
$f(x) = x^2 - 6x + 5$, domain $\mathbb{R}$
A table: $f(1) = 2,\; f(2) = 5,\; f(3) = 2$

Answer key

Yes — linear with nonzero slope is strictly increasing, so one-to-one.
No — vertex at $x = 3$. Restrict to $x \ge 3$ (or $x \le 3$).
Yes — strictly increasing on $[0, \infty)$.
No — vertex at $x = 3$. Restrict to $x \ge 3$ (or $x \le 3$).
No — the output $2$ is produced by both $1$ and $3$.

Your turn: give a quadratic and ask for the largest domain containing $x = 0$ on which it is invertible.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 10 / 16

Find an inverse of various functions algebraically, restricting the domain when necessary

The procedure

Write $y = f(x)$.
Swap $x$ and $y$.
Solve for $y$.
Write $y = f^{-1}(x)$, and state its domain (which equals the range of the original $f$).

Swapping first, then solving, keeps the algebra clean.

Worked example — linear

$f(x) = 3x - 7$. $y = 3x - 7 \;\to\; x = 3y - 7 \;\to\; y = \dfrac{x + 7}{3}$. So $f^{-1}(x) = \dfrac{x + 7}{3}$, domain $(-\infty, \infty)$.

Worked example — rational

$f(x) = \dfrac{x + 2}{x - 5}$.

\[ x = \frac{y + 2}{y - 5} \;\Rightarrow\; x(y - 5) = y + 2 \;\Rightarrow\; xy - 5x = y + 2 \] \[ xy - y = 5x + 2 \;\Rightarrow\; y(x - 1) = 5x + 2 \;\Rightarrow\; y = \frac{5x + 2}{x - 1} \]

$f^{-1}(x) = \dfrac{5x + 2}{x - 1}$, domain $(-\infty, 1) \cup (1, \infty)$ (the value $1$ is the horizontal asymptote of $f$ — an output $f$ never reaches).

Worked example — needs a domain restriction

$f(x) = x^2 - 4$ with domain restricted to $x \ge 0$.

\[ x = y^2 - 4 \;\Rightarrow\; y^2 = x + 4 \;\Rightarrow\; y = \pm\sqrt{x + 4} \]

The original domain was $x \ge 0$, so $f^{-1}$ must output nonnegative values → take the $+$ root. $f^{-1}(x) = \sqrt{x + 4}$, domain $[-4, \infty)$ (the range of $f$).

Watch out

$f^{-1}(x)$ is not $\dfrac{1}{f(x)}$.
When a $\pm$ appears, choose the sign using the original function's domain or range — don't keep both.
Always report the domain of $f^{-1}$. It is the range of $f$.
For a square-root function, the inverse is a restricted parabola, not the whole parabola.

Practice

Find $f^{-1}(x)$ and state its domain.

$f(x) = \dfrac{x}{4} + 2$
$f(x) = \dfrac{2x - 1}{x + 3}$
$f(x) = \sqrt{x - 1} + 2$, domain $x \ge 1$
$f(x) = (x + 5)^2$, domain $x \ge -5$

Answer key

$x = \tfrac{y}{4} + 2 \Rightarrow y = 4(x - 2) = 4x - 8$; domain $(-\infty, \infty)$.
$x(y + 3) = 2y - 1 \Rightarrow xy + 3x = 2y - 1 \Rightarrow y(x - 2) = -1 - 3x \Rightarrow y = \dfrac{-3x - 1}{x - 2}$; domain $(-\infty, 2) \cup (2, \infty)$.
$x - 2 = \sqrt{y - 1} \Rightarrow (x - 2)^2 = y - 1 \Rightarrow y = (x - 2)^2 + 1$; domain $[2, \infty)$.
$\sqrt{x} = y + 5 \Rightarrow y = \sqrt{x} - 5$ (positive root, since original $x \ge -5$ makes outputs $\ge -5$); domain $[0, \infty)$.

Your turn: give a rational function and ask for its inverse plus the excluded value in the inverse's domain.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 11 / 16

Use properties of functions to explain the relationship between a function and its inverse

What is always true of $f$ and $f^{-1}$

They undo each other: $f^{-1}(f(x)) = x$ for every $x$ in the domain of $f$, and $f(f^{-1}(x)) = x$ for every $x$ in the domain of $f^{-1}$. This is the definitive test that two functions are inverses.
Domain and range swap: $\text{domain}(f^{-1}) = \text{range}(f)$ and $\text{range}(f^{-1}) = \text{domain}(f)$.
Graphs are mirror images across $y = x$: if $(a, b)$ is on $f$, then $(b, a)$ is on $f^{-1}$.
Monotonicity is preserved: if $f$ is increasing, so is $f^{-1}$.
$x$- and $y$-intercepts trade places.

Worked example — verify by composition

$f(x) = 3x - 7$, $f^{-1}(x) = \dfrac{x + 7}{3}$.

\[ f^{-1}(f(x)) = \frac{(3x - 7) + 7}{3} = \frac{3x}{3} = x \quad\checkmark \] \[ f(f^{-1}(x)) = 3\!\left(\frac{x + 7}{3}\right) - 7 = (x + 7) - 7 = x \quad\checkmark \]

Point check: $f(4) = 5$ and $f^{-1}(5) = \dfrac{12}{3} = 4$. The point $(4, 5)$ on $f$ corresponds to $(5, 4)$ on $f^{-1}$.

Worked example — show two functions are NOT inverses

$f(x) = 2x + 3$, $g(x) = \dfrac{x}{2} - 3$.

\[ f(g(x)) = 2\!\left(\frac{x}{2} - 3\right) + 3 = x - 6 + 3 = x - 3 \neq x \]

Since one composition fails to return $x$, they are not inverses. (The correct inverse is $\dfrac{x - 3}{2}$.)

Watch out

You need both compositions to equal $x$ on the right domains — checking only one can mislead when domains are restricted.
The reflection is across the line $y = x$, not across the $x$-axis or $y$-axis.
If $f$ never reaches a value $c$ (a horizontal asymptote), then $c$ is excluded from the domain of $f^{-1}$.

Practice

Verify that $f(x) = \dfrac{x - 1}{4}$ and $g(x) = 4x + 1$ are inverses (both compositions).
$f$ is increasing with $f(0) = -3$ and $f(2) = 5$. Give two points on $f^{-1}$.
$f$ has domain $[0, \infty)$ and range $[-1, \infty)$. State the domain and range of $f^{-1}$.
Are $f(x) = x^2$ (on $x \ge 0$) and $g(x) = \sqrt{x}$ inverses? Check both compositions.

Answer key

$g(f(x)) = 4 \cdot \dfrac{x - 1}{4} + 1 = (x - 1) + 1 = x$; $f(g(x)) = \dfrac{(4x + 1) - 1}{4} = \dfrac{4x}{4} = x$. Inverses. ✓
$(-3, 0)$ and $(5, 2)$ — swap each coordinate pair.
Domain of $f^{-1}$ is $[-1, \infty)$; range of $f^{-1}$ is $[0, \infty)$.
$g(f(x)) = \sqrt{x^2} = |x| = x$ for $x \ge 0$; $f(g(x)) = (\sqrt{x})^2 = x$ for $x \ge 0$. Yes, inverses on those domains.

Your turn: give a pair of functions that pass one composition test but fail the other, and ask whether they're inverses.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Theme F

Rate of change, patterns, and piecewise models

The unit closes by turning functions toward data and the real world: how fast something changes, what pattern a table follows, and how to describe a rule that changes partway through.

Objective 12 / 16

Connect the rate of change function with a given scenario (distance/time, bottles, etc.)

The average rate of change of $f$ on $[a, b]$ is

\[ \frac{f(b) - f(a)}{b - a} \]

— the slope of the line through $\big(a, f(a)\big)$ and $\big(b, f(b)\big)$ (the secant line). Its units are output units per input unit.

Distance vs. time: the rate of change is speed. Steeper graph = faster; flat = stopped; straight line = constant speed.
Filling a bottle at a constant pour: graph water height vs. time. Where the bottle is narrow, height climbs fast (large rate of change); where it is wide, height climbs slowly. A bottle that bulges in the middle produces an S-shaped height–time curve.
The rate-of-change function itself: as the interval shrinks toward a single point, the average rate approaches the instantaneous rate — a new function telling you how fast $f$ is changing at each input. It is positive where $f$ increases, larger where $f$ is steeper, and zero at a peak or valley. (This is the bridge to calculus.)

Worked example

$h(t)$ is water height in cm after $t$ seconds, with $h(0) = 0$, $h(2) = 6$, $h(5) = 9$.

Average rate on $[0, 2]$: $\dfrac{6 - 0}{2 - 0} = 3$ cm/s.
Average rate on $[2, 5]$: $\dfrac{9 - 6}{5 - 2} = 1$ cm/s.
The level rose faster early, then slowed → the container is wider near the top.

Car example: if $d(1) = 40$ mi and $d(3) = 150$ mi, average speed on $[1, 3]$ is $\dfrac{150 - 40}{3 - 1} = 55$ mph.

Watch out

Always attach units to a rate of change.
Average rate over an interval is not the same as the function's value at one endpoint.
Sign matters: positive rate = increasing, negative = decreasing, zero = momentarily level.
"Narrow bottle → fast height increase" — many students get this backwards.

Practice

$f(x) = x^2$. Average rate of change on $[1, 4]$? On $[-2, 2]$?
A runner's position: $s(0) = 0$, $s(4) = 100$, $s(10) = 100$ (meters). Average speed on $[0,4]$? On $[4, 10]$? What happened after $t = 4$?
A cone (point down) is filled at a constant rate. Does the water height rise faster at the start or near the top? Why?
$P(t)$ doubles each hour. Is its rate-of-change function constant, increasing, or decreasing over time?

Answer key

$[1,4]$: $\dfrac{16 - 1}{3} = 5$. $[-2, 2]$: $\dfrac{4 - 4}{4} = 0$ (rises and falls symmetrically).
$[0,4]$: $25$ m/s. $[4, 10]$: $0$ m/s — the runner stopped (position unchanged).
Faster at the start: near the point the cone is narrow, so a fixed volume of water raises the level a lot; near the top it is wide, so the level rises slowly.
Increasing — each hour the amount added is larger, so the rate of change grows over time.

Your turn: give three container cross-sections and ask students to match each to a height–time graph.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 13 / 16

Identify patterns in given sets of data and connect to general functions

With inputs that are equally spaced (constant $\Delta x$), look at how the outputs change:

Pattern → function family

Constant first differences (each $y$ goes up by the same amount) → linear, $f(x) = mx + b$, with $m = \dfrac{\text{first difference}}{\Delta x}$.
Constant second differences → quadratic, $f(x) = ax^2 + bx + c$. With $\Delta x = 1$, $a = \dfrac{\text{second difference}}{2}$.
Constant ratio (each $y$ is a fixed multiple of the one before) → exponential, $f(x) = a \cdot b^{x}$, with $b = $ common ratio and $a = f(0)$.

Worked example — three tables

A. $x: 0,1,2,3,4$ $y: 5, 8, 11, 14, 17$. First differences all $3$ → linear. Slope $3$, $f(0) = 5$ → $f(x) = 3x + 5$.

B. $x: 0,1,2,3,4$ $y: 3, 4, 7, 12, 19$. First differences $1, 3, 5, 7$ (not constant); second differences $2, 2, 2$ → quadratic. $a = \tfrac{2}{2} = 1$. From $f(0) = 3$, $c = 3$; from $f(1) = 4 = 1 + b + 3$, $b = 0$. So $f(x) = x^2 + 3$ (check $f(3) = 12$ ✓).

C. $x: 0,1,2,3,4$ $y: 2, 6, 18, 54, 162$. Ratios all $\times 3$ → exponential. $a = f(0) = 2$, $b = 3$ → $f(x) = 2 \cdot 3^{x}$.

Watch out

Difference tables only work when inputs are equally spaced — check $\Delta x$ first.
Confirm the pattern across the whole table, not just the first two steps.
"Keeps getting bigger" is not enough to call it linear — linear needs a constant difference; exponential has a constant ratio.
A negative common ratio (e.g. $\times(-2)$) still signals exponential, with alternating signs.

Practice

Identify the family and find the rule.

$x: 0,1,2,3$ $y: 7, 4, 1, -2$
$x: 0,1,2,3,4$ $y: 1, 2, 5, 10, 17$
$x: 0,1,2,3$ $y: 100, 50, 25, 12.5$
$x: 1,2,3,4$ $y: 3, 12, 27, 48$

Answer key

First differences all $-3$ → linear: $f(x) = -3x + 7$.
First diffs $1,3,5,7$; second diffs $2$ → quadratic. $a = 1$, $c = 1$, $f(1) = 2 = 1 + b + 1 \Rightarrow b = 0$: $f(x) = x^2 + 1$.
Ratio $\times \tfrac{1}{2}$ → exponential: $f(x) = 100 \cdot \left(\tfrac{1}{2}\right)^{x}$.
$y = 3x^2$ — second differences constant at $6$ (note inputs start at $1$; $3(1)=3, 3(4)=12, 3(9)=27, 3(16)=48$).

Your turn: give a table that looks linear for the first two rows but turns out quadratic, and ask which family it is.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 14 / 16

Use function patterns to predict values

Once you've named the model (Objective 13), use it to fill gaps inside the data (interpolate) or reach beyond it (extrapolate). You can either plug into the rule, or continue the difference/ratio pattern step by step.

Worked examples

From a rule: Table B gave $f(x) = x^2 + 3$. Then $f(10) = 103$, and the output at $x = 7$ is $52$.
Continue a difference pattern: first differences are a constant $4$ and the last known point is $(6, 30)$. Next points: $(7, 34)$, $(8, 38)$, $(9, 42)$.
Continue a ratio pattern: constant ratio $\times 1.5$, last point $(4, 8)$. Next: $(5, 12)$, $(6, 18)$.
In context: bacteria start at $50$ and double every hour, so $P(t) = 50 \cdot 2^{t}$. After 6 h: $P(6) = 3200$. To reach $1600$: $50 \cdot 2^{t} = 1600 \Rightarrow 2^{t} = 32 \Rightarrow t = 5$ h.

Watch out

Extrapolation assumes the pattern keeps going. In a real scenario, say so.
Match the input to how $x$ is defined — "the year 2030" might be $x = 20$ if $x$ counts years since 2010.
For exponential models, "10% growth per year" means multiply by $1.10$, not add $0.10$.

Practice

A linear table has $f(0) = 12$ and constant first difference $-5$ (with $\Delta x = 1$). Find $f(4)$ and $f(10)$.
$f(x) = 2 \cdot 3^{x}$. Find $f(5)$; then find $x$ with $f(x) = 162$.
A quadratic table has second difference $4$ ($\Delta x = 1$), $f(0) = 1$, $f(1) = 3$. Find the rule, then $f(6)$.
A town of 8000 grows 5% per year. Population after 3 years? Write the model.

Answer key

$f(x) = 12 - 5x$: $f(4) = -8$, $f(10) = -38$.
$f(5) = 2 \cdot 243 = 486$. $2 \cdot 3^{x} = 162 \Rightarrow 3^{x} = 81 \Rightarrow x = 4$.
$a = \tfrac{4}{2} = 2$, $c = 1$; $f(1) = 2 + b + 1 = 3 \Rightarrow b = 0$. $f(x) = 2x^2 + 1$, so $f(6) = 73$.
$P(t) = 8000 \cdot 1.05^{t}$; $P(3) = 8000 \cdot 1.157625 \approx 9261$.

Your turn: give an exponential context and ask both a "predict the value" and a "when does it reach ___" question.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 15 / 16

Use piecewise functions to represent a graph

A piecewise function uses different rules on different parts of the domain:

\[ f(x) = \begin{cases} \text{rule}_1, & \text{condition}_1 \\ \text{rule}_2, & \text{condition}_2 \\ \ \vdots \end{cases} \]

Reading a graph into a formula

Split the $x$-axis at every corner, jump, or change of shape.
For each piece, find that segment's equation (line from slope + a point; parabola from its vertex; etc.).
Attach the interval. Decide $\le$ vs. $<$ from the dots: filled $\bullet$ = included, open $\circ$ = excluded. Every domain value should be covered exactly once.

To evaluate a piecewise function at $x = a$, first find which condition $a$ satisfies, then use only that piece.

Worked examples

Corner at the origin: the graph is $y = -x$ for $x < 0$ and $y = x^2$ for $x \ge 0$ (filled dot at the origin on the parabola):

\[ f(x) = \begin{cases} -x, & x < 0 \\ x^2, & x \ge 0 \end{cases} \]

Jump (step): height $1$ on $[0, 2)$ with an open dot at $x = 2$, then height $3$ on $[2, 4]$:

\[ f(x) = \begin{cases} 1, & 0 \le x < 2 \\ 3, & 2 \le x \le 4 \end{cases} \]

Two line pieces: from $(-3, 0)$ up to $(0, 3)$, then down to $(2, -1)$. Left slope $= \dfrac{3 - 0}{0 - (-3)} = 1$ through $(0,3)$: $y = x + 3$. Right slope $= \dfrac{-1 - 3}{2 - 0} = -2$ through $(0,3)$: $y = -2x + 3$.

\[ f(x) = \begin{cases} x + 3, & -3 \le x \le 0 \\ -2x + 3, & 0 < x \le 2 \end{cases} \]

Watch out

Conditions must not overlap — if both pieces claim $x = 2$, you've broken the "one output" rule. Line the inequalities up with the dots.
A jump between pieces means the graph is discontinuous there; that's allowed for a piecewise function.
Check an endpoint by plugging it into the piece whose condition includes it.
Don't leave a gap: every $x$ in the domain needs exactly one piece.

Practice

Write the piecewise rule: $y = 2$ (constant) for $x \le 1$, then the line through $(1, 2)$ and $(3, 6)$ for $x > 1$.
Evaluate that function at $x = 1$, $x = 0$, and $x = 3$.
A graph is made of three straight pieces: a ray coming down with slope $-1$ that arrives at $(-1, 0)$; a flat segment along the $x$-axis from $(-1, 0)$ to $(1, 0)$; and a ray leaving $(1, 0)$ with slope $1$. Write the piecewise rule, and decide whether the function is continuous.

Answer key

Right piece slope $= \dfrac{6 - 2}{3 - 1} = 2$, through $(1, 2)$: $y = 2x$. So $f(x) = \begin{cases} 2, & x \le 1 \\ 2x, & x > 1 \end{cases}$
$f(1) = 2$ (first piece); $f(0) = 2$; $f(3) = 6$.
Left ray through $(-1, 0)$ with slope $-1$: $y = -(x + 1) = -x - 1$. Right ray through $(1, 0)$ with slope $1$: $y = x - 1$. So $f(x) = \begin{cases} -x - 1, & x < -1 \\ 0, & -1 \le x \le 1 \\ x - 1, & x > 1 \end{cases}$ Check the joins: at $x = -1$, $-(-1) - 1 = 0$ equals the middle piece; at $x = 1$, $1 - 1 = 0$ equals the middle piece. The pieces meet with no gaps, so the function is continuous — though it has a corner at each breakpoint (the slope jumps), so it is not smooth.

Your turn: draw a graph with one corner and one jump, and ask for the piecewise rule with correct $\le$/$<$.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Objective 16 / 16

Use piecewise functions to represent and interpret a real-world scenario

Many real pricing and physical rules change partway through: tax brackets, shipping tiers, parking rates, phone-plan overages, overtime pay, utility rates. Each is naturally piecewise.

How to build one

Define the input variable and its units.
Find the breakpoints — the thresholds where the rule changes — and what each one means.
Write the rule for each piece (a flat fee is a constant; a per-unit rate is a slope).
Fix the boundary: does "40 or more" include exactly 40? Use $\ge$ or $>$ deliberately.
Interpret: evaluate at points of interest, and note whether the function is continuous at each breakpoint or jumps.

Worked example — overtime pay

\$18/hour for up to 40 hours, then time-and-a-half (\$27/hour) beyond 40. Let $h$ be hours worked.

\[ P(h) = \begin{cases} 18h, & 0 \le h \le 40 \\ 720 + 27(h - 40), & h > 40 \end{cases} \]

$P(45) = 720 + 27(5) = \$855$.
Continuous at $h = 40$: $18(40) = 720$ and $720 + 27(0) = 720$ agree. The graph bends upward — every hour past 40 is worth more.

Worked example — shipping (a jump)

\$6 flat shipping for orders under \$50; free for \$50 or more. Let $x$ be the order total in dollars.

\[ S(x) = \begin{cases} 6, & 0 \le x < 50 \\ 0, & x \ge 50 \end{cases} \]

$S(49.99) = \$6$, $S(50) = \$0$. The function jumps at $x = 50$ — spending one more cent saves \$6, which is exactly the "add an item to get free shipping" effect.

Watch out

Say what the input is and in what units before writing anything.
Get the boundary inequality right — "\$50 or more" includes $50$, so use $\ge$.
The second piece often needs a constant that carries the total from the breakpoint (the $720$ above), not a fresh start from zero.
Answer in a sentence with money/units, and say whether it's continuous or jumps if asked to interpret.

Practice

A garage charges \$4 for the first hour or any part of it, then \$2 for each additional hour, with a \$15 daily maximum. Sketch the idea and write $C(t)$ for $0 < t \le 3$ hours (whole-hour billing).
A phone plan: \$30 for up to 5 GB, then \$10 per additional GB. Write $B(g)$. Find $B(3)$ and $B(8)$.
For the overtime function $P(h)$ above, interpret $P(40) = 720$ and the change in slope at $h = 40$.
Is the phone-plan function $B(g)$ continuous at $g = 5$? Show it.

Answer key

$C(t) = \begin{cases} 4, & 0 < t \le 1 \\ 6, & 1 < t \le 2 \\ 8, & 2 < t \le 3 \end{cases}$ (each extra started hour adds \$2, up to the \$15 cap which isn't reached in 3 hours).
$B(g) = \begin{cases} 30, & 0 \le g \le 5 \\ 30 + 10(g - 5), & g > 5 \end{cases}$. $B(3) = \$30$; $B(8) = 30 + 10(3) = \$60$.
$P(40) = 720$: working exactly 40 hours pays \$720. At $h = 40$ the pay rate rises from \$18/h to \$27/h, so the graph's slope increases.
At $g = 5$: first piece gives $30$; second piece gives $30 + 10(0) = 30$. Equal → continuous.

Your turn: write a two-bracket tax scenario and ask for the piecewise rule, one evaluation, and whether it's continuous at the bracket.

✦ OnRamps Unit 1 Learning Guide — checkpoint

You have now covered of the 16 rows in the guide table:

Mark this row done on my Learning Guide — exam question written, confidence rated 1–3

↑ Contents

Lecture 2 · September 3, 2026

Exam-style practice problems

Problems in the format of the OnRamps Unit 1 quiz, with full solutions. Nothing new here — each one is an application of an objective you have already met. Reveal a solution only after you have committed to an answer.

Which objective each problem leans on

Q1 — is $y$ a function of $x$? → Objective 1
Q3 — domain from a formula → Objective 4
Q4 — the definition of an even function → Objective 8
Q5 — increasing / decreasing intervals → Objective 3
Q6 — restrict a parabola's domain, then invert it → Objectives 9–10
Q7 — even / odd / neither from a graph → Objective 8

Fast strategies

Function or not, from an equation: solve for $y$. A single clean expression is a function; a $\pm$ (from $y^2$ or an even root) means one $x$ gives two $y$'s — not a function.
Function or not, from a table: a repeated input only breaks it if its outputs disagree.
Domain from a formula: start with all reals; force even roots $\ge 0$; forbid denominators $= 0$; then combine.
Even / odd from a graph: fold across the $y$-axis → matches = even; rotate $180^\circ$ about the origin → matches = odd; otherwise neither. A line with a nonzero $y$-intercept is neither.
Invertible parabola: one-to-one only on one side of the vertex. For $a(x - h)^2 + k$ the vertex is at $x = h$; restrict to $x \ge h$ or $x \le h$.
Find $f^{-1}$: swap $x$ and $y$, solve; when $\pm$ appears, keep the sign that lands inside the restricted domain.

Question 1 · 18 pts

Select whether each relationship represents $y$ as a function of $x$

$5x + 4y = 7$
$x = y^2 - 16$
$x^2 + y^2 = 36$
The table $x:\ -1,\ 2,\ 0,\ 1,\ 2$ $y:\ 4,\ -2,\ 1,\ 5,\ -2$

Solution

Show solution

Function. Solve for $y$: $4y = 7 - 5x \Rightarrow y = \dfrac{7 - 5x}{4}$. Every $x$ gives exactly one $y$.
Not a function. $y^2 = x + 16 \Rightarrow y = \pm\sqrt{x + 16}$. At $x = 0$, $y = \pm 4$ — one input, two outputs.
Not a function. $y = \pm\sqrt{36 - x^2}$: a circle of radius $6$, which fails the vertical line test ($x = 0 \Rightarrow y = \pm 6$).
Function. The input $2$ appears twice, but both times the output is $-2$, so no input has two different outputs. (A repeated input is fine when its outputs agree.)

Question 3 · domain matching

Give the domain of each function

The quiz gave three functions and asked you to match each to its domain. The answer statements were "all real values less than or equal to 5," "all real values except $-6$ and $8$," and "all real values greater than or equal to 5 except 7." A function of each type:

$g(x) = \sqrt{5 - x}$
$h(x) = \dfrac{2x}{(x + 6)(x - 8)}$
$k(x) = \dfrac{\sqrt{x - 5}}{x - 7}$

Solution

Show solution

Need $5 - x \ge 0 \Rightarrow x \le 5$. Domain: all real values $\le 5$ $= (-\infty, 5\,]$.
The denominator is zero at $x = -6$ and $x = 8$; nothing else is restricted. Domain: all reals except $-6$ and $8$ $= (-\infty, -6) \cup (-6, 8) \cup (8, \infty)$.
The square root needs $x - 5 \ge 0$ ($x \ge 5$); the denominator forbids $x = 7$. Domain: all real values $\ge 5$ except $7$ $= [5, 7) \cup (7, \infty)$.

Question 4 · 5 pts

True or false: an even function is a function where $f(-x) = f(x)$

Solution

Show solution

True. That equation is the definition of an even function — the output is unchanged when the input's sign flips, so the graph is symmetric across the $y$-axis. (An odd function is the other one: $f(-x) = -f(x)$, symmetric about the origin.)

Question 5 · 12 pts

Is the graph increasing or decreasing on the given intervals?

Representative version: the graph below rises to a peak at $(-2, 7)$, falls to a valley at $(1, -3)$, then rises again. State where it is increasing and where it is decreasing.

Peak at $(-2, 7)$, valley at $(1, -3)$.

Solution

Show solution

Read left to right — up to the peak, down to the valley, then up again:

Increasing on $(-\infty, -2) \cup (1, \infty)$.
Decreasing on $(-2, 1)$.

Give $x$-intervals as open intervals; the peak and valley belong to neither.

Question 6 · 12 pts

For $f(x) = 3(x + 2)^2 - 4$: restrict the domain to make it invertible, then find $f^{-1}$

Answer choices for $f^{-1}(x)$:

$\sqrt{\dfrac{x - 4}{3}} - 2$
$\dfrac{\sqrt{x + 4}}{3} + 2$
$\dfrac{\sqrt{x - 4}}{3} + 2$
$\sqrt{\dfrac{x + 4}{3}} - 2$

Solution

Show solution

Domain restriction. The vertex is at $x = -2$. A parabola is one-to-one only on one side of its vertex, so restrict to $x \ge -2$ (or $x \le -2$). The listed inverse matches the $x \ge -2$ branch.

Invert. Swap $x$ and $y$, then solve:

\[ x = 3(y + 2)^2 - 4 \;\Rightarrow\; \frac{x + 4}{3} = (y + 2)^2 \;\Rightarrow\; y + 2 = \pm\sqrt{\frac{x + 4}{3}} \]

The restricted domain is $x \ge -2$, so the outputs of $f^{-1}$ must be $\ge -2$: take the $+$ root.

\[ f^{-1}(x) = \sqrt{\frac{x + 4}{3}} - 2 \qquad \textbf{(choice D)} \]

Its domain is $[-4, \infty)$, the range of $f$. Choices B and C put the $3$ outside the root — it belongs inside, dividing $x + 4$; choice A uses $x - 4$ instead of $x + 4$.

Question 7 · 21 pts

Identify each graph as even, odd, or neither

Classify each graph by symmetry: even = mirror image across the $y$-axis; odd = unchanged by a $180^\circ$ turn about the origin; neither = no such symmetry.

(a)

(b)

(c)

(d)

(e)

(f)

Solution

Show solution

(a) even. Upper semicircle centred on the $y$-axis ($y = \sqrt{r^2 - x^2}$); it folds exactly onto itself across the $y$-axis.
(b) odd. An $x^3$-type curve through the origin; a $180^\circ$ turn about the origin leaves it unchanged.
(c) even. Horizontal line $y = c$: $f(-x) = c = f(x)$.
(d) odd. A $\sin$-type wave through the origin — origin symmetry.
(e) neither. A straight line that does not pass through the origin: no $y$-axis or origin symmetry. (A line through the origin, $y = mx$, would be odd.)
(f) odd. $y = x^3 - 3x$: a local max and a local min placed symmetrically about the origin — $f(-x) = -f(x)$.

Algebra check when you have the formula: compute $f(-x)$. Equal to $f(x)$ → even; equal to $-f(x)$ → odd; neither → neither. And if a graph fails the vertical line test (a sideways parabola, say), it isn't a function, so "even / odd" doesn't apply.

↑ Contents

Lecture 3 · September 8, 2026

Piecewise functions & rates of change

Worked solutions to the Discovery Precalculus explorations on piecewise functions (1.5.3) and on rates — distance–time graphs (1.3.1) and filling bottles (1.3.2) — plus a DeltaMath graphing task and two OnRamps quiz questions on identifying function patterns. These build on Objectives 12–16.

The two ideas in this lecture

Piecewise: different rules on different parts of the domain. To read one from a graph, split the $x$-axis at every corner or jump, find each segment's equation, then attach the interval — deciding $\le$ vs. $<$ from the filled (•) and open (○) dots. To evaluate at a number, use only the piece whose condition that number satisfies.
Rate of change: on a distance–time graph the slope is speed; on a height–volume graph the slope is set by how wide the container is right there — narrow means the level climbs fast (steep), wide means it climbs slowly (shallow).

Exploration 1.5.3 · #1 · read a formula from a graph

Write a piecewise formula for the graph

Read the three pieces off the graph — a parabola, a constant segment, then a line — using the filled (•) and open (○) dots to fix each endpoint.

Solution

Show solution

Left piece: a parabola with vertex at the origin, ending with a solid dot at $(1, 1)$ → $y = x^2$ for $x \le 1$.

Middle piece: a horizontal segment at height $3$, open at $(1, 3)$ and solid at $(2, 3)$ → $y = 3$ for $1 < x \le 2$.

Right piece: a line of slope $1$ that starts at an open dot at $(2, 2)$ and continues → $y = x$ for $x > 2$.

\[ f(x) = \begin{cases} x^2, & x \le 1 \\ 3, & 1 < x \le 2 \\ x, & x > 2 \end{cases} \]

Two jumps: at $x = 1$ the value leaps from $1$ up to $3$; at $x = 2$ it drops from $3$ down to $2$. The domain is all real numbers.

Exploration 1.5.3 · #2 · sketch & classify

Sketch the relation, and decide whether it is a function

\[ y = \begin{cases} x + 3, & x \le -1 \\ x^2 - 1, & -1 < x \le 2 \\ 1, & x \ge 2 \end{cases} \]

Solution

Show solution

It is not a function. Look at $x = 2$: it satisfies both $-1 < x \le 2$ (giving $x^2 - 1 = 3$) and $x \ge 2$ (giving $1$). One input, two outputs — a vertical line at $x = 2$ hits the graph at both $(2, 3)$ and $(2, 1)$.

The two stacked points at $x = 2$ are why it fails the vertical line test.

The fix a textbook usually intends: change the last condition to $x > 2$. Then every input has exactly one output and it is a function (with a jump at $x = -1$ and another at $x = 2$).

Exploration 1.5.3 · #3 · evaluate & graph

Evaluate $f(-1), f(0), f(2), f(4), f(6)$, then graph

(a) $f(x) = \begin{cases} x^2 - 2, & x < 2 \\ 4 + x, & x \ge 2 \end{cases}$ (b) $f(x) = \begin{cases} 4 - x^2, & x < 1 \\ \sqrt{x + 1}, & x \ge 1 \end{cases}$

Solution

Show solution

(a) $x = -1, 0$ use $x^2 - 2$; $x = 2, 4, 6$ use $4 + x$.

$f(-1) = 1 - 2 = -1$, $f(0) = -2$, $f(2) = 6$, $f(4) = 8$, $f(6) = 10$.

(a) — jump from $2$ up to $6$ at $x = 2$.

(b) $x = -1, 0$ use $4 - x^2$; $x = 2, 4, 6$ use $\sqrt{x + 1}$.

$f(-1) = 4 - 1 = 3$, $f(0) = 4$, $f(2) = \sqrt{3}$, $f(4) = \sqrt{5}$, $f(6) = \sqrt{7}$.

(b) — jump down from $3$ to $\sqrt{2}$ at $x = 1$.

Exploration 1.5.3 · #4 · build a model

Gunterville College tuition as a piecewise function

\$92 per credit for the first 10 credits; \$35 per credit for credits 11–18; \$74 per credit beyond 18. Write total tuition $T$ as a function of the number of credits $c$.

Solution

Show solution

Add the cost bracket by bracket (like tax brackets). The first 10 credits cost $10 \times 92 = \$920$; credits 11–18 add up to $8 \times 35 = \$280$, for \$1200 at 18 credits.

\[ T(c) = \begin{cases} 92c, & 0 \le c \le 10 \\ 920 + 35(c - 10), & 10 < c \le 18 \\ 1200 + 74(c - 18), & c > 18 \end{cases} \]

Check the seams: at $c = 10$, $92(10) = 920$ and $920 + 35(0) = 920$; at $c = 18$, $920 + 35(8) = 1200$ and $1200 + 74(0) = 1200$. Continuous, and piecewise-linear with slopes $92 \to 35 \to 74$ (a bulk discount in the middle band, a surcharge for overload).

Reasonable domain and range: $c$ is a whole number of credits, $0 \le c \le 24$ or so (a realistic overload cap). $T$ then takes the discrete values $T(0) = \$0$ up to $T(24) = 1200 + 74(6) = \$1644$.

Exploration 1.5.3 · #5 · evaluate & graph

Data-plan cost: find $C(0.5)$ and $C(6)$, then graph

\[ C(g) = \begin{cases} 25, & 0 < g < 2 \\ 25 + 10(g - 2), & g \ge 2 \end{cases} \]

Solution

Show solution

$0.5$ is in $0 < g < 2$, so $C(0.5) = \$25$ — the flat rate.

$6 \ge 2$, so $C(6) = 25 + 10(6 - 2) = 25 + 40 = \$65$.

Flat \$25 until $g = 2$, then \$10 per extra GB. The marked points are $C(0.5) = 25$ and $C(6) = 65$.

The two pieces agree at $g = 2$ (both give \$25), so the graph has no jump — just a corner where the price starts rising.

DeltaMath · graph a piecewise function

Graph the piecewise function on the axes

\[ f(x) = \begin{cases} 2x + 5, & -3 \le x < 0 \\ -x + 4, & 0 < x < 6 \end{cases} \]

Solution

Show solution

Each piece is a line — plot its two endpoints and join them, using the inequality to decide filled vs. open.

$2x + 5$ on $[-3, 0)$: from $(-3, -1)$ filled up to $(0, 5)$ open.
$-x + 4$ on $(0, 6)$: from $(0, 4)$ open down to $(6, -2)$ open.

Note $x = 0$: the first condition is $x < 0$ and the second is $x > 0$, so $0$ is in neither — $f(0)$ is undefined, and both endpoints at $x = 0$ are open circles.

Exploration 1.3.1 · distance–time graphs

Describe the motion that produces each distance–time graph

Time is on the horizontal axis, distance from a fixed reference on the vertical axis. The slope is the velocity, so the shape of the graph translates directly into a description of the walk. Eight of the textbook's graphs are redrawn below (numbered as in the book).

#1

#4

#5

#6

#7

#8

#9

#12

Reading guide & answers

Show solution

Graph	Motion
#1 — falling line then rising line	Walk toward the detector at a constant speed, then away at a constant speed.
#4 — up, flat, down	Walk away at a constant speed, stand still (function constant), then walk back at a constant speed.
#5 — triangle (up to a peak, then down)	Walk away at a constant speed, turn around at the far point, walk back at a constant speed.
#6 — curve getting steeper	Walk away and speed up — increasing distance at an increasing rate.
#7 — curve flattening toward a horizontal	Walk away but slow down, easing to a stop — increasing distance at a decreasing rate.
#8 — decreasing curve flattening near the axis	Walk toward the detector, slowing down as you get close.
#9 — smooth arch (∩)	Walk away — speeding up then slowing — pause at the farthest point, then walk back the same way. A smooth version of #5.
#12 — staircase with vertical jumps	Not a function. A vertical piece means two distances at one instant — physically impossible, so no walk produces it.

A full write-up names each interval, the direction of travel, whether the speed is constant / faster / slower, and any interval where the walker is stopped.

Exploration 1.3.2 · bottles, rates & graphs

Sketch height vs. volume as each bottle is filled at a steady rate

Pour in liquid at a constant rate and plot the height $h$ of the liquid against the volume $v$ added. Where the bottle is narrow, a little volume raises the level a lot → the graph is steep. Where the bottle is wide, the level barely moves → the graph is shallow. The five bottles and their graphs:

Reading guide & answers

Show solution

Bottle	Height–volume graph
a · sphere	Steep near the bottom, flattening through the wide middle, then steepening again near the top — an S-shape.
b · cylinder (constant width)	A straight line — the level rises at a constant rate.
c · hourglass (wide, pinched, wide)	Shallow while the wide base fills, steep through the narrow pinch, then shallow again up the wide top.
d · stepped (bands of different width)	Straight within each band, with the slope jumping at every step — steeper where the band is narrow, shallower where it is wide.
e · funnel (wide top, narrow bottom)	Starts very steep where the bottle is narrow, then curves to ever-shallower as it widens — concave down.

The graph is always increasing (adding liquid never lowers the level); only its steepness changes — smoothly when the bottle's width changes smoothly, abruptly at a step.

OnRamps quiz · classify the pattern

Is each table linear, logarithmic, or exponential?

The sum-and-product chart

Look at how the list of $x$-values is built, then how the list of $y$-values is built. A list built by adding the same amount each time is a sum list; one built by multiplying by the same amount each time is a product list.

$x$-list	$y$-list	function
sum ($+c$)	sum ($+d$)	linear
sum ($+c$)	product ($\times r$)	exponential
product ($\times k$)	sum ($+d$)	logarithmic
product ($\times k$)	product ($\times r$)	power

If the table's $x$-values aren't in a clean chain, pick a few that are — e.g. $2, 4, 8$ or $10, 20, 40$ form a product chain — and check what their $y$-values do.

$x$	$y$
4	2
8	6
12	18
16	54

$x$	$y$
10	6
20	13
30	17
40	20

$x$	$y$
2	12
6	18
12	21
18	27

Solution

Show solution

Table 1 — exponential. $x$: $4, 8, 12, 16$ is a sum list ($+4$ each time). $y$: $2, 6, 18, 54$ is a product list ($\times 3$ each time). Sum-$x$ with product-$y$ → exponential.

Table 2 — logarithmic. The $x$-values $10, 20, 40$ form a product chain ($\times 2$). Their $y$-values $6, 13, 20$ form a sum chain ($+7$). Product-$x$ with sum-$y$ → logarithmic. (Reading the full table the other way: $x$ goes up by $10$ but $y$ climbs by less and less — $7, 4, 3$ — the same tell.)

Table 3 — power. The $x$-values $2, 6, 18$ form a product chain ($\times 3$). Their $y$-values $12, 18, 27$ form a product chain ($\times 1.5$). Product-$x$ with product-$y$ → power. (The remaining point $x = 12, y = 21$ fits this only roughly, and if the choices are just linear / log / exponential the points sit closest to a line — which is the answer marked on the quiz.)

OnRamps quiz · Question 5 · 12 pts

Given $f(2) = 3$ and $f(4) = 9$, find $f(8)$ under four models

No equations to solve — use the sum-and-product chart from the previous problem. The inputs $2, 4, 8$ form a product chain ($\times 2$), and the one $y$-step you are given, $3 \to 9$, is both $+6$ and $\times 3$. Each model says what the next $y$-step must be; just continue the pattern.

Solution

Show solution

a. Linear — equal additions to $x$ give equal additions to $y$. Step $x$ by $2$: $2 \to 4 \to 6 \to 8$. The step $x{:}\,2\to4$ took $y$ from $3$ to $9$, i.e. $+6$. So $y$: $3,\ 9,\ 15,\ \mathbf{21}$.

b. Exponential — equal additions to $x$ give equal multiples of $y$. Step $x$ by $2$: $2 \to 4 \to 6 \to 8$. The step $2\to4$ multiplied $y$ by $3$ ($3 \to 9$). So $y$: $3,\ 9,\ 27,\ \mathbf{81}$.

c. Power — multiplying $x$ by a constant multiplies $y$ by a constant. Multiply $x$ by $2$: $2 \to 4 \to 8$. The step $2\to4$ multiplied $y$ by $3$ ($3 \to 9$). So $y$: $3,\ 9,\ \mathbf{27}$.

d. Logarithmic — multiplying $x$ by a constant adds a constant to $y$. Multiply $x$ by $2$: $2 \to 4 \to 8$. The step $2\to4$ added $6$ to $y$ ($3 \to 9$). So $y$: $3,\ 9,\ \mathbf{15}$.

Model	$x$-step used	what it does to $y$	$f(8)$
Linear	$+2$	$+6$	$\mathbf{21}$
Exponential	$+2$	$\times 3$	$\mathbf{81}$
Power	$\times 2$	$\times 3$	$\mathbf{27}$
Logarithmic	$\times 2$	$+6$	$\mathbf{15}$

↑ Contents

Lecture 4 · September 21, 2026

Transformations of graphs

Three OnRamps quiz questions about transforming a parent function: writing an equation from the graph of a square-root curve (two versions of Question 4), writing an absolute-value equation in the quiz's $\pm G\,\text{function}(x-H)+K$ form (Question 6), and matching a transformed graph to its formula (Question 8). Read the short lesson first, then try each problem before opening its solution.

The one idea: every move has a job

Start from a parent function — a simple graph you already know, like $\sqrt{x}$ or $|x|$. Every equation in this lecture is that parent with four adjustments:

\[ f(x) \;=\; \pm\,G\;\text{function}(x - H) \;+\; K \]

$H$ slides the graph right by $H$ (so $x + 4$ means $H = -4$: left 4). It sits inside the function, so it does the opposite of what the sign suggests.
$K$ slides the graph up by $K$. It sits outside, so it does exactly what it says.
$G$ stretches the graph vertically by a factor of $G$ (compresses it if $0 < G < 1$).
$\pm$ a minus sign flips the graph upside-down over the $x$-axis.

To move a single point of the parent, $(a, b)$, to its new home: $\;(a, b) \;\longmapsto\; (a + H,\ \pm G\,b + K)$.

Reading an equation off a graph — three steps

Find the start. A square root starts at an endpoint; an absolute value turns at a corner (the vertex). That point is $(H, K)$.
Take one "nice" step from it. For $\sqrt{x}$, the parent goes right 1, up 1; for $|x|$, also right 1, up 1. How far the real graph actually moves up or down over that one step is $\pm G$ — down means the minus sign.
Check with another point. Substitute it into your equation. If it fails, redo step 2.

Two parent points worth memorizing: $\sqrt{x}$ passes through $(0,0),\ (1,1),\ (4,2),\ (9,3)$; $|x|$ passes through $(0,0),\ (1,1),\ (2,2)$ and $(-1,1),\ (-2,2)$.

Changes inside the parentheses (Question 8)

Write the transformed function as $h(x) = \pm f\big(b(x - c)\big) + d$. Then, for every point $(u, v)$ on the graph of $f$, the new point is

\[ (u, v) \;\longmapsto\; \Big(\dfrac{u}{b} + c,\ \ \pm v + d\Big). \]

Outside the $f$ ($\pm$ and $+d$) acts on the $y$-values, as written.
Inside acts on the $x$-values, backwards: the inside $\tfrac{1}{2}(x+1)$ means "add 1, then halve," so the graph is doubled wider, then moved left 1. Multiplying inside by $b$ divides the widths by $b$: $b = \tfrac12$ stretches $\times 2$; $b = 2$ squeezes to half.

Shortcut that never fails: set the inside equal to the old $x$-value and solve for the new $x$. For $\tfrac12(x+1) = u$: $x = 2u - 1$.

OnRamps quiz · Question 4 · 6 pts · version 1

Write an equation for the graph below as a transformation of the square root function

The curve starts at the marked point on the left and passes through the other three marked points.

Solution

Show solution

Step 1 — the start. The curve begins at $(-4,\ 2)$, so $H = -4$ and $K = 2$. The inside of the root is $x - (-4) = x + 4$.

Step 2 — one nice step. One unit right of the start is $x = -3$, where the point $(-3,\ -1)$ sits. The parent $\sqrt{x}$ would go up 1 over that step; this curve goes from $y = 2$ down to $y = -1$, a drop of $3$. So there is a flip (minus sign) and $G = 3$.

Step 3 — check with the other points:

$x = 0$: $-3\sqrt{0+4} + 2 = -3(2) + 2 = -4$ ✓ (the point $(0,-4)$)
$x = 5$: $-3\sqrt{5+4} + 2 = -3(3) + 2 = -7$ ✓ (the point $(5,-7)$)

\[ \boxed{\,f(x) = -3\sqrt{x + 4} + 2\,} \]

Domain $[-4, \infty)$; range $(-\infty, 2]$ — the graph never rises above its starting height.

OnRamps quiz · Question 4 · 6 pts · version 2 (with a submitted answer)

Write an equation for the graph below as a transformation of the square root function

Same question, different graph. The screenshot shows a submitted answer, not yet graded: $-2\sqrt{x+3}+4$. Work it out yourself first — then see whether the submitted answer holds up.

Solution

Show solution

Step 1 — the start. The curve begins at $(-3,\ 4)$: $H = -3$, $K = 4$, so the root contains $x + 3$.

Step 2 — one nice step. One unit right, at $x = -2$, the curve is at $(-2,\ 2)$: down $2$ from height $4$. So there is a minus sign and $G = 2$.

Step 3 — check:

$x = 1$: $-2\sqrt{1+3} + 4 = -2(2) + 4 = 0$ ✓ (the point $(1, 0)$)
$x = 6$: $-2\sqrt{6+3} + 4 = -2(3) + 4 = -2$ ✓ (the point $(6, -2)$)

\[ \boxed{\,f(x) = -2\sqrt{x + 3} + 4\,} \]

The submitted answer is correct — it matches all four marked points.

Domain $[-3, \infty)$; range $(-\infty, 4]$.

Watch out

Inside signs are backwards. A start at $x = -3$ gives $\sqrt{x + 3}$, not $\sqrt{x - 3}$. Test with the start: the inside must be $0$ there.
Do not use the total drop as $G$ unless you stepped exactly 1 right. For $\sqrt{x}$, one step right lifts the parent by exactly 1 — that is why the drop over one step is $G$. Over 4 steps the parent lifts by 2, so a drop over four steps would be $2G$.

OnRamps quiz · Question 6 · 15 pts

Write an equation for the function graphed below

Fill in $f(x) = \pm G\,\text{function}(x - H) + K$: choose the sign, $G$, the parent function, $H$, and $K$. The corner is at $(2, 3)$ and the graph also passes through $(5, 1)$.

Solution

Show solution

Which parent? Two straight arms meeting in a sharp corner — a "V" shape (here upside-down): the absolute value function, $|x|$.

The corner is $(H, K)$. The corner is at $(2, 3)$, so $H = 2$ and $K = 3$.

The sign. The corner is the highest point — the arms go down from it — so the graph is flipped: the sign is −.

$G$. From the corner to $(5, 1)$: right $3$, down $2$. The parent $|x|$ rises exactly as much as it moves right, so the steepness here is $\tfrac{2}{3}$ of that: $G = \tfrac{2}{3}$.

Check. $-\tfrac23|5-2| + 3 = -\tfrac23(3) + 3 = 1$ ✓. On the left arm at $x = 0$: $-\tfrac23|0-2| + 3 = -\tfrac43 + 3 = \tfrac53 \approx 1.7$ — the graph crosses the $y$-axis a little below 2 ✓.

Dropdown	Answer
$\pm$	$-$
$G$	$\tfrac{2}{3}$
function	$\|x\|$ (absolute value)
$H$	$2$
$K$	$3$

\[ \boxed{\,f(x) = -\tfrac{2}{3}\,|x - 2| + 3\,} \]

Domain $(-\infty, \infty)$; range $(-\infty, 3]$. The graph crosses the $x$-axis where $\tfrac23|x-2| = 3$, i.e. $|x-2| = 4.5$: at $x = -2.5$ and $x = 6.5$.

OnRamps quiz · Question 8 · 6 pts

Given the graph of $f(x)$, which of the following is the graph of $h(x) = -f\!\left(\tfrac12(x+1)\right) + 2$?

Here is $f$. It starts at $(-4, 1)$, rises along a smooth arch to $(-3, 2)$, drops to a sharp point at $(-1, -2)$, then a straight line climbs through the origin to $(2, 4)$.

The six answer choices (redrawn from the quiz screenshots; they are the same size and scale, $x$ from $-10$ to $6$):

A

B

C

D

E

F

Solution

Show solution

Strategy: track four key points. Use the ends and corners of $f$: $(-4, 1),\ (-3, 2),\ (-1, -2),\ (2, 4)$.

Vertical part (outside): $-f(\dots) + 2$ — flip the $y$-value, then add 2. Horizontal part (inside): $\tfrac12(x+1) = u$ gives $x = 2u - 1$ — double the old $x$, then subtract 1.

point on $f$ $(u, v)$	new $x = 2u - 1$	new $y = -v + 2$	point on $h$
$(-4, 1)$	$-9$	$-1 + 2 = 1$	$(-9, 1)$
$(-3, 2)$	$-7$	$-2 + 2 = 0$	$(-7, 0)$
$(-1, -2)$	$-3$	$2 + 2 = 4$	$(-3, 4)$
$(2, 4)$	$3$	$-4 + 2 = -2$	$(3, -2)$

So $h$ starts at $(-9, 1)$, dips to a smooth low point at $(-7, 0)$ (the old top of the arch, flipped), climbs to a sharp peak at $(-3, 4)$ (the old sharp bottom, flipped), then falls in a straight line to $(3, -2)$.

Choose by elimination — each check throws out a wrong choice:

The flip (minus sign). $f$ has an arch on top of the left side; $h$ must have a valley there, and a peak where $f$ had its sharp bottom. That removes A and F, which keep the arch upward.
Stretch, not squeeze. The inside factor $\tfrac12$ makes the graph twice as wide ($f$ spans $-4$ to $2$, 6 wide; $h$ spans 12 wide). C and E are only about half as wide as $f$ — they squeezed instead. Eliminated.
Which way it moved. Between the two that remain, the sharp peak must be at $x = -3$ ($-1$ doubled is $-2$, minus 1). D has its peak at $x = -1$ — it moved right 1 instead of left 1. Eliminated.

Only B is left, and it matches all four key points: $(-9,1),\ (-7,0),\ (-3,4),\ (3,-2)$. Answer: B. (This is the choice marked on the screenshot, scored 6 / 6.)

Optional check with formulas. The arch of $f$ is $2 - (x+3)^2$ and the line is $2x$. Substituting $\tfrac12(x+1)$ for $x$, flipping and adding 2 gives $h(x) = \tfrac{(x+7)^2}{4}$ for $-9 \le x \le -3$, and $h(x) = 1 - x$ for $-3 \le x \le 3$ — a parabola opening upward with low point $(-7, 0)$, then a line of slope $-1$.

Watch out

Factor inside = the opposite of what you expect. $\tfrac12$ inside is a stretch by 2, and $2$ inside is a squeeze to half. A quick sanity check: a factor of $\tfrac12$ means the function needs to be fed twice the $x$ to do the same thing, so it is slower — wider.
Do the two directions separately. The $y$-changes (flip, $+2$) never affect the $x$-coordinates and vice-versa. Fill a table like the one above and choose from the corners.
Sign of the shift. $(x + 1)$ moves the graph left — but because of the $\tfrac12$, the shift is applied after the stretch: new $x = 2u - 1$, not $2(u - 1)$.

↑ Contents

Lecture 5 · September 28, 2026

Graphing ellipses

A new shape: the ellipse — a stretched circle. Given an equation, find its center, vertices, co-vertices, and foci, and sketch it; given a graph, write its equation. This starts the conic-sections part of the course and does not map onto the Unit 1 Learning Guide.

Standard form

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]

The center is $(h, k)$. The two denominators, $a^2$ and $b^2$, tell you how far the ellipse reaches from the center along each axis — always take $a$ and $b$ to be positive, and always take $a$ to be the larger of the two (so $a \ge b$). Whichever variable, $x$ or $y$, sits under $a^2$ tells you which way the ellipse is long:

$a^2$ under $x$ → major axis horizontal. Vertices (the far ends) are $(h \pm a,\ k)$; co-vertices (the near ends) are $(h,\ k \pm b)$; foci are $(h \pm c,\ k)$.
$a^2$ under $y$ → major axis vertical. Vertices are $(h,\ k \pm a)$; co-vertices are $(h \pm b,\ k)$; foci are $(h,\ k \pm c)$.

Either way, the two foci sit on the long axis, a distance $c$ from the center, where

\[ c^2 = a^2 - b^2. \]

Anatomy of an ellipse

(this example is horizontal: $a$ is under $x$)

Reading an equation — four steps

Center. Read $(h, k)$ straight from $x - h$ and $y - k$ — watch the sign: $(x+2)$ means $h = -2$.
$a$ and $b$. Square-root each denominator. Whichever is bigger is $a$; the other is $b$.
Vertices & co-vertices. Move $a$ from the center along the axis whose denominator held $a^2$; move $b$ along the other axis.
Foci. Compute $c = \sqrt{a^2 - b^2}$, then move $c$ from the center along the same axis as the vertices (the long axis).

Reading a graph, or getting to standard form first

From a picture: find the center (the middle of the oval), then count boxes from the center to the left/right edge and to the top/bottom edge. The larger count is $a$ — and it tells you which axis is major.
From a messy equation like $64x^2 + 25y^2 = 1600$ or $36(x-2)^2 + 4(y+3)^2 - 144 = 0$: move any lone number to the other side, then divide every term by whatever makes the right side equal $1$.

Question 1 · identify & graph

$\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ — find the center, $a$, $b$, $c$, vertices, co-vertices, and foci

Solution

Show solution

Center: $(0, 0)$ — no $h$ or $k$ shown means both are $0$.

$9 > 4$ and $9$ sits under $x$, so the major axis is horizontal: $a = \sqrt{9} = 3$, $b = \sqrt{4} = 2$.

$c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5}$.

Center	$(0,0)$
$a,\ b,\ c$	$a = 3,\ \ b = 2,\ \ c = \sqrt{5}$
Vertices	$(3, 0)$ and $(-3, 0)$
Co-vertices	$(0, 2)$ and $(0, -2)$
Foci	$(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$

To sketch: plot the center, go $3$ left/right for the vertices and $2$ up/down for the co-vertices, then draw a smooth oval through all four points. The foci sit inside the ellipse on the long axis — they are not on the curve itself.

Question 2 · identify & graph

$\dfrac{(x+2)^2}{25} + \dfrac{(y-1)^2}{9} = 1$ — find the center, $a$, $b$, $c$, vertices, co-vertices, and foci

Solution

Show solution

Center: $(x+2)$ means $h = -2$; $(y - 1)$ means $k = 1$. Center $(-2, 1)$.

$25 > 9$ and $25$ sits under $x$, so the major axis is horizontal: $a = \sqrt{25} = 5$, $b = \sqrt{9} = 3$.

$c = \sqrt{25 - 9} = \sqrt{16} = 4$.

Center	$(-2, 1)$
$a,\ b,\ c$	$a = 5,\ \ b = 3,\ \ c = 4$
Vertices	$(-2 + 5,\ 1) = (3, 1)$ and $(-2 - 5,\ 1) = (-7, 1)$
Co-vertices	$(-2,\ 1+3) = (-2, 4)$ and $(-2,\ 1-3) = (-2, -2)$
Foci	$(-2+4,\ 1) = (2, 1)$ and $(-2-4,\ 1) = (-6, 1)$

Same method as Question 1, just starting from $(-2, 1)$ instead of the origin — every point above is the origin-centered answer shifted left 2 and up 1.

Watch out

The sign flips. $(x + 2)$ in the equation means the center's $x$-coordinate is $-2$, not $+2$ — same rule as with parabolas and absolute value graphs.
$a$ is whichever is bigger, not whichever is under $x$. If the larger number were under $y$ instead, the ellipse would be vertical and the vertices/foci would move up-down, not left-right.
$c^2 = a^2 - b^2$, never $a^2 + b^2$. The foci are always inside the ellipse, closer to the center than the vertices are ($c < a$).

Question 5 · standard form

Write $64x^2 + 25y^2 = 1600$ in standard form

Solution

Show solution

The right side needs to be $1$, so divide every term by $1600$:

\[ \frac{64x^2}{1600} + \frac{25y^2}{1600} = \frac{1600}{1600} \]

Simplify each fraction ($1600 \div 64 = 25$, $1600 \div 25 = 64$):

\[ \boxed{\,\frac{x^2}{25} + \frac{y^2}{64} = 1\,} \]

Here $64 > 25$ and $64$ is under $y$, so this ellipse is vertical: center $(0,0)$, $a = 8$, $b = 5$, $c = \sqrt{64-25} = \sqrt{39}$. Vertices $(0, \pm 8)$; co-vertices $(\pm 5, 0)$; foci $(0, \pm\sqrt{39})$.

Question 6 · standard form

Write $36(x-2)^2 + 4(y+3)^2 - 144 = 0$ in standard form

Solution

Show solution

Move the $-144$ to the other side:

\[ 36(x-2)^2 + 4(y+3)^2 = 144 \]

Divide every term by $144$ ($144 \div 36 = 4$, $144 \div 4 = 36$):

\[ \boxed{\,\frac{(x-2)^2}{4} + \frac{(y+3)^2}{36} = 1\,} \]

Center $(2, -3)$. Here $36 > 4$ and $36$ is under $y$, so this ellipse is vertical: $a = 6$, $b = 2$, $c = \sqrt{36-4} = \sqrt{32} = 4\sqrt{2}$. Vertices $(2, 3)$ and $(2, -9)$; co-vertices $(4, -3)$ and $(0, -3)$; foci $(2,\ -3+4\sqrt{2})$ and $(2,\ -3-4\sqrt{2})$.

Watch out

Divide every single term — including the one that's already just a number on the right. It's easy to divide the $x$- and $y$-terms but forget to also divide $144$.
Simplify the fractions so each denominator is a perfect square when possible — that's what lets you read $a$ and $b$ off directly instead of leaving messy roots.

Question 7 · graph → equation

Given the graph below, write the equation of the ellipse in standard form

Read the center and the reach of the ellipse straight off the grid.

Solution

Show solution

Center. The oval runs from $x = -5$ to $x = -1$ (middle: $-3$) and from $y = -4$ to $y = 4$ (middle: $0$). Center $(-3, 0)$.

Reach. Horizontally the ellipse reaches $2$ boxes from center ($-3$ to $-1$, or $-3$ to $-5$). Vertically it reaches $4$ boxes from center ($0$ to $4$, or $0$ to $-4$).

$4 > 2$ and the larger reach is vertical, so this ellipse is tall: $a = 4$ goes with $y$, $b = 2$ goes with $x$.

\[ \boxed{\,\frac{(x+3)^2}{4} + \frac{y^2}{16} = 1\,} \]

Check: $c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \approx 3.46$, so the foci sit at $(-3,\ 2\sqrt3)$ and $(-3,\ -2\sqrt3)$ — inside the ellipse, above and below center, matching the tall shape.

↑ Contents

\(x\)-list	\(y\)-list	function
sum (\(+c\))	sum (\(+d\))	linear
sum (\(+c\))	product (\(\times r\))	exponential
product (\(\times k\))	sum (\(+d\))	logarithmic
product (\(\times k\))	product (\(\times r\))	power

Model	\(x\)-step used	what it does to \(y\)	\(f(8)\)
Linear	\(+2\)	\(+6\)	\(\mathbf{21}\)
Exponential	\(+2\)	\(\times 3\)	\(\mathbf{81}\)
Power	\(\times 2\)	\(\times 3\)	\(\mathbf{27}\)
Logarithmic	\(\times 2\)	\(+6\)	\(\mathbf{15}\)

Dropdown	Answer
\(\pm\)	\(-\)
\(G\)	\(\tfrac{2}{3}\)
function	\(\|x\|\) (absolute value)
\(H\)	\(2\)
\(K\)	\(3\)

point on \(f\) \((u, v)\)	new \(x = 2u - 1\)	new \(y = -v + 2\)	point on \(h\)
\((-4, 1)\)	\(-9\)	\(-1 + 2 = 1\)	\((-9, 1)\)
\((-3, 2)\)	\(-7\)	\(-2 + 2 = 0\)	\((-7, 0)\)
\((-1, -2)\)	\(-3\)	\(2 + 2 = 4\)	\((-3, 4)\)
\((2, 4)\)	\(3\)	\(-4 + 2 = -2\)	\((3, -2)\)

Center	\((0,0)\)
\(a,\ b,\ c\)	\(a = 3,\ \ b = 2,\ \ c = \sqrt{5}\)
Vertices	\((3, 0)\) and \((-3, 0)\)
Co-vertices	\((0, 2)\) and \((0, -2)\)
Foci	\((\sqrt{5}, 0)\) and \((-\sqrt{5}, 0)\)

Center	\((-2, 1)\)
\(a,\ b,\ c\)	\(a = 5,\ \ b = 3,\ \ c = 4\)
Vertices	\((-2 + 5,\ 1) = (3, 1)\) and \((-2 - 5,\ 1) = (-7, 1)\)
Co-vertices	\((-2,\ 1+3) = (-2, 4)\) and \((-2,\ 1-3) = (-2, -2)\)
Foci	\((-2+4,\ 1) = (2, 1)\) and \((-2-4,\ 1) = (-6, 1)\)

\(x\)	\(y\)
4	2
8	6
12	18
16	54

\(x\)	\(y\)
10	6
20	13
30	17
40	20

\(x\)	\(y\)
2	12
6	18
12	21
18	27

Foundations of functions

Use multiple representations to identify functions and justify what defines a function

Apply function notation to evaluate a function at a value or expression

Behavior, domain, and meaning

Determine increasing and decreasing intervals of various functions

Identify the domain and range of various functions using interval, set, and/or inequality notation

Determine function values and inverse values and connect their meaning with a given scenario or model

Composition of functions

Determine the composite function from a given pair of functions and their associated domain

Evaluate composite functions at a constant, taking the domain into account

Properties and classification

Identify function properties (even, odd, neither) and types of functions (injective, surjective, bijective) from graphs, mappings, tables, equations, etc.

Symmetry: even, odd, or neither

Type: injective, surjective, bijective

Inverse functions

Predict when a function will have an inverse

Find an inverse of various functions algebraically, restricting the domain when necessary

Use properties of functions to explain the relationship between a function and its inverse

Rate of change, patterns, and piecewise models

Connect the rate of change function with a given scenario (distance/time, bottles, etc.)

Identify patterns in given sets of data and connect to general functions

Use function patterns to predict values

Use piecewise functions to represent a graph

Use piecewise functions to represent and interpret a real-world scenario

Exam-style practice problems

Select whether each relationship represents \(y\) as a function of \(x\)

Give the domain of each function

True or false: an even function is a function where \(f(-x) = f(x)\)

Is the graph increasing or decreasing on the given intervals?

For \(f(x) = 3(x + 2)^2 - 4\): restrict the domain to make it invertible, then find \(f^{-1}\)

Identify each graph as even, odd, or neither

Piecewise functions & rates of change

Write a piecewise formula for the graph

Sketch the relation, and decide whether it is a function

Evaluate \(f(-1), f(0), f(2), f(4), f(6)\), then graph

Gunterville College tuition as a piecewise function

Data-plan cost: find \(C(0.5)\) and \(C(6)\), then graph

Graph the piecewise function on the axes

Describe the motion that produces each distance–time graph

Sketch height vs. volume as each bottle is filled at a steady rate

Is each table linear, logarithmic, or exponential?

Given \(f(2) = 3\) and \(f(4) = 9\), find \(f(8)\) under four models

Transformations of graphs

Write an equation for the graph below as a transformation of the square root function

Write an equation for the graph below as a transformation of the square root function

Write an equation for the function graphed below

Given the graph of \(f(x)\), which of the following is the graph of \(h(x) = -f\!\left(\tfrac12(x+1)\right) + 2\)?

Graphing ellipses

\(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\) — find the center, \(a\), \(b\), \(c\), vertices, co-vertices, and foci

\(\dfrac{(x+2)^2}{25} + \dfrac{(y-1)^2}{9} = 1\) — find the center, \(a\), \(b\), \(c\), vertices, co-vertices, and foci

Write \(64x^2 + 25y^2 = 1600\) in standard form

Write \(36(x-2)^2 + 4(y+3)^2 - 144 = 0\) in standard form

Given the graph below, write the equation of the ellipse in standard form

Before the Unit 1 exam