Algebra 3 Quiz: Function Transformations

12 questions · exam conditions

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Function TransformationsQuestion 1 of 12

Starting from $f(x)=x^2$ , which sequence of transformations, applied in the order listed, produces $h(x)=-\frac{1}{2}(x-4)^2+3$ ?

Reflect across the

x

-axis, compress vertically by a factor of

\frac{1}{2}

, shift 4 units right, shift 3 units up.

Reflect across the

x

-axis, stretch vertically by a factor of 2, shift 4 units left, shift 3 units up.

Compress vertically by a factor of

\frac{1}{2}

, shift 4 units right, shift 3 units up, reflect across the

x

-axis.

Shift 4 units right, shift 3 units up, reflect across the

x

-axis, compress vertically by a factor of

\frac{1}{2}

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Algebra 3 Quiz

Algebra 3 Quiz: Function Transformations

Practice Function Transformations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Function Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Starting from $f(x)=x^2$ , which sequence of transformations, applied in the order listed, produces $h(x)=-\frac{1}{2}(x-4)^2+3$ ?

Reflect across the $x$ -axis, compress vertically by a factor of $\frac{1}{2}$ , shift 4 units right, shift 3 units up. (correct answer)
Reflect across the $x$ -axis, stretch vertically by a factor of 2, shift 4 units left, shift 3 units up.
Compress vertically by a factor of $\frac{1}{2}$ , shift 4 units right, shift 3 units up, reflect across the $x$ -axis.
Shift 4 units right, shift 3 units up, reflect across the $x$ -axis, compress vertically by a factor of $\frac{1}{2}$ .

Explanation: Whenever you see a transformed quadratic like

h(x)=-\frac12(x-4)^2+3

, start from

f(x)=x^2

and compare to the form

y=a(x-h)^2+k

. Here

a=-\frac12

h=4

, and

k=3

. The negative sign means reflect across the

x

-axis, the factor

\frac12

means compress vertically,

(x-4)

means shift 4 units right, and

+3

means shift 3 units up. Applying the vertical changes first, then the shifts, gives exactly: reflect to

-x^2

, compress to

-\frac12 x^2

, shift right to

-\frac12(x-4)^2

, then shift up to

-\frac12(x-4)^2+3

. The distractor that says "stretch vertically by a factor of 2, shift 4 units left" misreads the reciprocal and confuses the horizontal direction. The sequence "compress, shift right, shift up, then reflect" would produce

-\left(\frac12(x-4)^2+3\right)=-\frac12(x-4)^2-3

, turning the final shift downward. The sequence "shift right, shift up, reflect, then compress" gives

-\frac12(x-4)^2-\frac32

, because the reflection and compression are applied to the vertical shift too. Order truly matters. A good strategy: first identify the reflection/compression coefficient and the shifts from the form

y=a(x-h)^2+k

, then apply vertical changes before shifts unless you are careful about how later transformations affect earlier shifts.

Question 2

The graph of $g$ is a transformation of $f(x)=|x|$ . Its vertex is $(2,-3)$ , it opens downward, and it passes through $(3,-5)$ . Which equation represents $g$ ?

$g(x)=-2|x-2|-3$ (correct answer)
$g(x)=-2|x+2|-3$
$g(x)=2|x-2|-3$
$g(x)=-2|x-2|+3$

Explanation: Whenever you see a transformed absolute value graph, write it in vertex form:

g(x)=a|x-h|+k

, where

(h,k)

is the vertex, and

a

controls direction and stretch. The vertex is

(2,-3)

, so you need

|x-2|

and an outside

-3

. That already points to

g(x)=-2|x-2|-3

. It opens downward, so

a

must be negative; that rules out

g(x)=2|x-2|-3

. It also passes through

(3,-5)

: substituting gives

-5=a|3-2|-3

, so

a=-2

. Perfect. Now the distractors. The choice

g(x)=-2|x+2|-3

has the same shape and vertical shift, but

|x+2|

means

x-(-2)

, so its vertex is

(-2,-3)

, not

(2,-3)

. The choice

g(x)=2|x-2|-3

has the correct vertex, but a positive

a

makes it open upward. Finally,

g(x)=-2|x-2|+3

has the correct horizontal shift and downward opening, but the outside

+3

puts the vertex at

(2,3)

, not

(2,-3)

. A quick strategy: identify the vertex first from the signs inside and outside the absolute value, then use the direction to check the sign of

a

, and use a given point to find its exact value. Remember that

|x-2|

shifts right, while

|x+2|

shifts left — this sign confusion is a classic trap.

Question 3

Which transformation of $f(x)=2^x$ produces the same graph as $g(x)=2^{x+3}$ ?

A reflection across the $x$ -axis followed by a vertical shift up 3 units
A vertical shift up 3 units
A horizontal shift right 3 units
A vertical stretch by a factor of 8 (correct answer)

Explanation: Whenever you see a transformation question, rewrite the target in terms of

f(x)

and decide whether the change is inside the exponent or outside it. Here,

g(x)=2^{x+3}=2^x\cdot 2^3=8\cdot 2^x=8f(x).

That means every output value of

f

is multiplied by

8

, so the graph is exactly the vertical stretch of

f

by a factor of

8

. The other choices confuse where the

+3

acts. A reflection across the

x

-axis followed by a vertical shift up

3

would produce

-2^x+3

, not

8\cdot2^x

. A vertical shift up

3

would produce

2^x+3

, which adds

3

to each output instead of multiplying it by

8

. A horizontal shift right

3

would produce

f(x-3)=2^{x-3}=2^x\cdot2^{-3}=\frac{2^x}{8},

which compresses the graph vertically by

8

, not stretches it. The equivalent horizontal move would actually be a shift left

3

, since

f(x+3)=2^{x+3}

, but that choice is not listed. The key habit: distinguish input changes (inside the exponent, horizontal shifts) from output changes (outside the exponent, vertical shifts and stretches). Then use exponent laws to convert forms—you will quickly see that

g

is just

f

stretched vertically by

8

Question 4

Let $f(x)=\sqrt{x}$ . Which function represents $f$ stretched horizontally by a factor of 3, then shifted down 2 units?

$g(x)=\sqrt{x+3}-2$
$g(x)=\sqrt{3x}-2$
$g(x)=\sqrt{\frac{x}{3}}-2$ (correct answer)
$g(x)=3\sqrt{x}-2$

Explanation: When you see function transformations, ask yourself: is the change inside the function (affecting

x

) or outside it (affecting the output)? Inside changes control horizontal behavior; outside changes control vertical behavior. And remember horizontal transformations work inversely to intuition. . To stretch

f(x)=\sqrt{x}

horizontally by a factor of 3, you replace

x

with

\frac{x}{3}

, giving

\sqrt{\frac{x}{3}}

. A horizontal stretch pulls points away from the

y

-axis, so dividing

x

by the factor is correct. Then shifting down 2 units means subtracting 2 from the whole function:

g(x)=\sqrt{\frac{x}{3}}-2

This is the correct choice. Why the others are traps?

$\sqrt{x+3}-2$ shifts the graph left 3 units because adding inside moves the graph in the negative $x$ -direction — a translation, not a stretch.
$\sqrt{3x}-2$ multiplies $x$ by 3 inside, which actually compresses horizontally by a factor of $\frac13$ , not stretches by 3. This is the classic inverse-effect trap.
$3\sqrt{x}-2$ multiplies the output by 3, a vertical stretch, not a horizontal one. It changes height, not width.

A quick memory aid: Inside is opposite. To stretch horizontally by factor

a

, divide

x

a

. To shift vertically, do the normal thing outside — subtract 2 to move down. On exam day, rewrite the transformations step-by-step and check whether the

x

is divided or multiplied.

Question 5

Starting with $f(x)=\sqrt{x}$ , reflect its graph across the $x$ -axis and then shift it 4 units up. What are the domain and range of the transformed function?

Domain: $x\ge 0$ ; range: $y\ge 4$
Domain: $x\ge 0$ ; range: $y\le 4$ (correct answer)
Domain: $x\le 0$ ; range: $y\le 4$
Domain: $x\ge 0$ ; range: $y\le -4$

Explanation: When you see a transformation question, identify each operation in order and ask: does this change the inputs (

x

) or the outputs (

y

)? Reflecting across the

x

-axis and shifting vertically affect only outputs, so the domain of the square root function stays

x \ge 0

. Start with

f(x)=\sqrt{x}

, where

\sqrt{x}\ge 0

. Reflect across the

x

-axis:

y=-\sqrt{x}

, so outputs become

\le 0

. Then shift 4 units up:

y=-\sqrt{x}+4

. Now every output is

4

minus a nonnegative number, so the largest possible output is

4

, and outputs decrease without bound below it. Thus the range is

y \le 4

. The correct answer is domain

x\ge 0

; range

y\le 4

. Each wrong choice traps a specific misunderstanding. "Domain:

x\ge 0

; range:

y\ge 4

" treats the upward shift as raising the lower bound, but a negative square root flips the inequality direction. "Domain:

x\le 0

; range:

y\le 4

" incorrectly assumes reflecting across the

x

-axis somehow changes the domain; reflections across the

x

-axis do not affect input values. "Domain:

x\ge 0

; range:

y\le -4

" confuses the 4-unit shift with moving the range below

-4

; the shift is upward, so the ceiling should be

+4

, not

-4

. Study tip: for transformations, remember that vertical stretches/reflections/shifts change the range, while horizontal ones change the domain. Track the graph of

y=\sqrt{x}

through each step, and write the final equation before finding domain and range.

Question 6

The graph of $f$ contains the point $(1,4)$ . The graph of $g(x)=f(2x-6)$ is a transformation of the graph of $f$ . Which point must be on the graph of $g$ ?

$(0.5,4)$
$(3.5,4)$ (correct answer)
$(5,4)$
$(1,4)$

Explanation: Whenever you see a transformation question like

g(x)=f(2x-6)

, focus on what happens to the input of

f

. Since

(1,4)

is on

f

, you know

f(1)=4

. For

g

to output

4

, you need the inside expression to equal

1

: set

2x-6=1

. Solving gives

2x=7

, so

x=3.5

. Therefore

(3.5,4)

must be on

g

. This is the correct point. The other choices show common missteps. The point

(5,4)

comes from solving

x-6=1

after forgetting the coefficient

2

, or from shifting right by

6

instead of applying the stretch. The point

(0.5,4)

might come from solving

2x=1

and ignoring the

-6

, treating the transformation as just a horizontal compression. The point

(1,4)

assumes that the original point passes through unchanged, which would only be true if

g(x)=f(x)

or if

2x-6=1

x=1

, but

2(1)-6=-4

, not

1

. A reliable strategy: rewrite

2x-6

2(x-3)

, but then set the whole inside equal to the known input, not just part of it. For any point

(a,b)

f

, find

x

such that the expression inside

f

equals

a

. This turns transformations into a simple equation and avoids visual guessing. On the algebra exam, this input-equation method works for all horizontal stretches, compressions, and shifts.

Question 7

Starting with the graph of $y=f(x)$ , compress it horizontally by a factor of $\frac{1}{2}$ , then shift the result 2 units to the right. Which equation describes the resulting graph?

$y=f(2x-2)$
$y=f(2x-4)$ (correct answer)
$y=f(2x)-2$
$y=f(x-2)$

Explanation: Whenever you see horizontal compressions or shifts, remember that all horizontal changes happen inside

f

's parentheses—and they work in the opposite direction from what you might expect: compress by

\frac12

means multiply

x

2

, and shift right

2

means replace

x

x-2

. Start with

y=f(x)

. Compress horizontally by a factor of

\frac12

: inside becomes

2x

, so you get

y=f(2x)

. Now shift that graph 2 units to the right: replace every

x

f(2x)

with

x-2

. That gives

y=f(2(x-2))=f(2x-4).

That is the correct choice. The choice

y=f(2x-2)

is a common trap: it factors as

f(2(x-1))

, so it compresses by

\frac12

but then shifts right only 1 unit, not 2. The choice

y=f(2x)-2

correctly compresses, but the

-2

is outside the function, making it a vertical shift down 2, not a horizontal shift right. The choice

y=f(x-2)

shifts the graph 2 units right but never applies the horizontal compression. A good check: after compressing, factor out the coefficient before reading the shift. In

f(2x-4)=f(2(x-2))

, the shift right is clearly 2. Horizontal transformations are inside the parentheses; vertical shifts are outside. Keep those two worlds separate.

Question 8

Let $g(x)=(2x)^3-1$ . How is the graph of $g$ related to the graph of $f(x)=x^3$ ?

It is compressed vertically by a factor of $\frac{1}{2}$ and shifted down 1 unit.
It is stretched horizontally by a factor of 2 and shifted down 1 unit.
It is compressed horizontally by a factor of $\frac{1}{2}$ and shifted down 1 unit. (correct answer)
It is stretched horizontally by a factor of 2 and shifted up 1 unit.

Explanation: When you see a transformation question like this, the key is to distinguish changes inside the parentheses from changes outside. Inside the function affects horizontal behavior; outside affects vertical behavior. Here,

g(x)=(2x)^3-1

. Starting from

f(x)=x^3

, the

2

inside the parentheses multiplies every

x

before cubing. That compresses the graph horizontally by a factor of

\frac{1}{2}

: points move closer to the

y

-axis. Then the

-1

outside shifts the entire graph down 1 unit. So the correct relationship is a horizontal compression by

\frac{1}{2}

and a downward shift of 1. The other choices trap common confusions. "Compressed vertically by a factor of

\frac{1}{2}

" mixes up axes: since

(2x)^3=8x^3

, the actual vertical change is a stretch by 8, not a compression. "Stretched horizontally by a factor of 2" reverses the effect: multiplying by 2 compresses, not stretches; a horizontal stretch by 2 would use

\left(\frac{x}{2}\right)^3

. Finally, "shifted up 1 unit" misreads the sign: the constant term is

-1

, so the graph moves down, not up. On exam day, remember: transformations inside the parentheses are horizontal and often "opposite" of intuition, while outside changes are vertical and follow the sign directly. Always test a point, like

x=1

, to check direction and scale.

Question 9

The function $g$ is formed by taking $f(x)=\frac{1}{x}$ , shifting its graph 2 units left, reflecting it across the $x$ -axis, and shifting it 1 unit down. Which statement about the graph of $g$ is true?

The vertical asymptote is $x=-2$ , and the horizontal asymptote is $y=1$ .
The vertical asymptote is $x=2$ , and the horizontal asymptote is $y=-1$ .
The vertical asymptote is $x=-2$ , and the horizontal asymptote is $y=-1$ . (correct answer)
The vertical asymptote is $x=2$ , and the horizontal asymptote is $y=1$ .

Explanation: When you see a transformation question involving a rational function like

f(x)=\frac1x

, track each shift and reflection one step at a time. The original vertical asymptote is

x=0

, and the horizontal asymptote is

y=0

. Shifting the graph 2 units left replaces

x

with

x+2

, giving

y=\frac1{x+2}

, so the vertical asymptote moves to

x=-2

. Reflecting across the

x

-axis gives

y=-\frac1{x+2}

. Finally, shifting 1 unit down gives

g(x)=-\frac1{x+2}-1

. Because

-\frac1{x+2}\to 0

x\to\pm\infty

, the horizontal asymptote is

y=-1

. So the true statement is: vertical asymptote

x=-2

, horizontal asymptote

y=-1

. The choice saying the vertical asymptote is

x=2

reverses the direction of the shift — moving left means subtracting from

x

, which lands on

-2

, not

2

. The choice saying the horizontal asymptote is

y=1

misses the combined effect of the reflection and downward shift: the reflection flips the sign, and the shift moves it down, so the asymptote goes to

-1

, not

1

. The choice with both

x=2

and

y=1

combines both errors, and the choice with

x=-2

but

y=1

correctly identifies the vertical shift direction but not the final vertical position. On exam day, write the transformed equation first. The asymptotes come directly from the denominator's zero and the end behavior — don't rely on memory of the graph alone.

Question 10

Suppose $f(3)=7$ . Let $g(x)=-2f(x-1)+5$ . Based only on this information, which point must lie on the graph of $g$ ?

$(2,-9)$
$(3,-9)$
$(4,19)$
$(4,-9)$ (correct answer)

Explanation: This question tests function transformations: horizontal shifts happen inside the parentheses, while vertical stretches and shifts happen outside. When you see a "must lie on the graph" question with a known point like

f(3)=7

, your goal is to find the

x

-value that makes the inside expression equal to

3

. Here, set

x-1=3

, so

x=4

. Then

g(4)=-2f(3)+5=-2(7)+5=-14+5=-9

. Therefore, the point

(4,-9)

must lie on the graph of

g

. The choice

(2,-9)

comes from incorrectly treating

x-1

as the input value itself, or shifting in the wrong direction. The choice

(3,-9)

uses

3

as the new input, but

g(3)

would require

f(2)

, which is unknown. The choice

(4,19)

correctly finds

x=4

but makes a sign error: it computes

+14+5

instead of

-14+5

. In general, be careful that multiplying by

-2

changes both the magnitude and the sign. A strong strategy is to start from the known point

(3,7)

and apply the transformations in order: shift right by

1

to get

(4,7)

, reflect and stretch vertically by

-2

to get

(4,-14)

, then shift up

5

to get

(4,-9)

. Matching "inside equals known input" first will always locate the guaranteed point.

Question 11

The graph of $g$ is obtained from $f(x)=\sqrt{x}$ by shifting 3 units left, vertically stretching by a factor of 2, and then reflecting across the $x$ -axis. Which of the following is $g(x)$ ?

$g(x)=-2\sqrt{x+3}$ (correct answer)
$g(x)=2\sqrt{-x+3}$
$g(x)=-2\sqrt{x-3}$
$g(x)=2\sqrt{x+3}$

Explanation: Whenever you see a function transformation problem, separate horizontal changes inside the function from vertical changes outside. For

f(x)=\sqrt{x}

, shifting 3 units left means replacing

x

x+3

, giving

\sqrt{x+3}

. A vertical stretch by factor 2 multiplies the output by 2:

2\sqrt{x+3}

. Then reflecting across the

x

-axis multiplies the output by

-1

, so

g(x)=-2\sqrt{x+3}

. The choice

2\sqrt{-x+3}

has a different horizontal structure:

-x+3=3-x

, which reflects the graph across the

y

-axis and shifts it right, not left; it also lacks the

x

-axis reflection. The choice

-2\sqrt{x-3}

has the vertical stretch and

x

-axis reflection, but

x-3

shifts the graph right 3 units, not left. The choice

2\sqrt{x+3}

has the correct left shift and vertical stretch, but multiplying by 2 instead of

-2

means the

x

-axis reflection is missing. A useful habit is to build the transformations step-by-step and check the sign inside the radical:

x+

constant means shift left,

x-

constant means shift right. A negative sign outside the function is your signal that the final reflection across the

x

-axis has been applied.

Question 12

Starting with the graph of $y=f(x)$ , shift the graph 3 units to the right, then reflect it across the $y$ -axis. Which equation describes the resulting graph?

$y=f(-x)+3$
$y=f(-x+3)$
$y=f(x-3)$
$y=f(-x-3)$ (correct answer)

Explanation: Whenever a transformations question appears, focus on what happens inside the function versus outside. Inside changes affect

x

horizontally; outside changes affect

y

vertically. Also, horizontal shifts are counterintuitive: shifting right 3 means replacing

x

with

x-3

, so the shifted graph is

y=f(x-3)

. Now reflect across the

y

-axis: reflection replaces every

x

with

-x

. Apply that to the already-shifted expression:

y=f((-x)-3)=f(-x-3)

. So the resulting graph is indeed

y=f(-x-3)

. Now see why the others miss the mark.

y=f(x-3)

only describes the shift right; it forgets the reflection entirely.

y=f(-x)+3

is a vertical shift upward by 3, not a horizontal shift or a true input reflection; it confuses inside and outside transformations. The tempting choice

y=f(-x+3)

often comes from reflecting first and then shifting right, or shifting left and then reflecting—either way, the order of operations is wrong. Since the problem first shifts right, then reflects, the

-3

must stay inside after the reflection, giving

-x-3

, not

-x+3

. Study tip: track transformations step by step, rewriting the function after each move. For horizontal changes, ask yourself exactly what

x

is replaced by before moving on. This prevents the classic "reflection order" trap.