Algebra 2 Quiz: Transformations Of Functions And Graphs

20 questions · exam conditions

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Transformations Of Functions And GraphsQuestion 1 of 20

How does replacing $f(x)$ with $f(3x)$ change the graph of $f(x)$ ?

It shifts the graph right 3 units

It compresses the graph horizontally by a factor of 3 (all

x

-values are divided by 3)

It stretches the graph horizontally by a factor of 3 (all

x

-values are multiplied by 3)

It stretches the graph vertically by a factor of 3 (all

y

-values are multiplied by 3)

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Algebra 2 Quiz

Algebra 2 Quiz: Transformations Of Functions And Graphs

Practice Transformations Of Functions And Graphs in Algebra 2 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 Transformations Of Functions And Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

How does replacing $f(x)$ with $f(3x)$ change the graph of $f(x)$ ?

It shifts the graph right 3 units
It compresses the graph horizontally by a factor of 3 (all $x$ -values are divided by 3) (correct answer)
It stretches the graph horizontally by a factor of 3 (all $x$ -values are multiplied by 3)
It stretches the graph vertically by a factor of 3 (all $y$ -values are multiplied by 3)

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Replacing with f(3x) affects inside the function, so it's horizontal: since |3| > 1, it compresses horizontally by a factor of 3, meaning x-values are scaled by 1/3 or divided by 3 to squeeze the graph toward the y-axis. Choice B correctly identifies the transformation as compressing horizontally by a factor of 3 with x-values divided by 3. A distractor like D might confuse it with vertical effects, but remember, the '3' is inside with x, so it's horizontal, not vertical! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 2

Let the parent function be $f(x)=x^2$ , whose graph is the standard parabola with vertex at $(0,0)$ . What transformation changes $f(x)$ to $g(x)=2f(x-3)+1$ ?

Shift left 3 units, stretch vertically by a factor of 2, then shift down 1 unit
Shift right 3 units, compress vertically by a factor of $\tfrac{1}{2}$ , then shift up 1 unit
Shift right 3 units, stretch vertically by a factor of 2, then shift up 1 unit (correct answer)
Shift left 3 units, compress horizontally by a factor of 2, then shift up 1 unit

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. For g(x) = 2f(x-3) + 1, start with f(x-3) which shifts the parabola right by 3 units (since it's f(x - 3), making x larger to achieve the same output), then multiply by 2 to stretch vertically by a factor of 2, and finally add 1 to shift up 1 unit. Choice C correctly identifies the transformation as shifting right 3 units, stretching vertically by a factor of 2, then shifting up 1 unit. A common mistake, like in choice A, is thinking f(x-3) shifts left instead of right—remember, the shift direction is opposite the sign inside the function! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 3

Let $f(x)=x^2$ . The graph of $g$ is shown on the coordinate plane along with $f$ . The function $g$ is of the form $g(x)=f(x+k).$ What is the value of $k$ ?

(Use the graph: $f$ has vertex at $(0,0)$ and $g$ has vertex at $(2,0)$ .)

$k=2$
$k=-2$ (correct answer)
$k=\tfrac{1}{2}$
$k=-\tfrac{1}{2}$

Explanation: This question tests your understanding of how algebraic transformations of functions—like

f(x) + k

k \cdot f(x)

f(kx)

, and

f(x + k)

—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1)

f(x) + k

shifts the graph vertically (up if

k > 0

, down if

k < 0

), (2)

k \cdot f(x)

stretches vertically if

|k| > 1

or compresses if

0 < |k| < 1

(and reflects across x-axis if

k < 0

), (3)

f(x + k)

shifts horizontally—LEFT if

k > 0

, RIGHT if

k < 0

(opposite of what you might expect!), (4)

f(kx)

compresses horizontally if

|k| > 1

or stretches if

0 < |k| < 1

(and reflects across y-axis if

k < 0

). Outside the function (

f(x) + k

and

k \cdot f(x)

) affects y-values/vertical; inside the function (

f(x + k)

and

f(kx)

) affects x-values/horizontal. Given

g(x) = f(x + k)

with

f(x) = x^2

and the graph showing g's vertex at

(2,0)

compared to f's at

(0,0)

, this indicates a horizontal shift right by 2 units, so

f(x + k) = (x + k)^2

has vertex at

(-k, 0)

; setting

-k = 2

gives

k = -2

. Choice B correctly finds the k value as

-2

. An error like in choice A might ignore the opposite sign rule for horizontal shifts—

f(x + 2)

would shift left to

(-2,0)

, not right, so k must be negative for a right shift. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like

f(x) + 3

2f(x)

) changes y-values (vertical effects). Anything done INSIDE the parentheses (like

f(x + 3)

f(2x)

) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—

f(x + 3)

shifts LEFT 3 because you're subtracting 3 from x-coordinates, and

f(x - 2)

shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 4

Determine whether the function $f(x)=x^4-2x^2$ is even, odd, or neither.

Neither, because polynomials cannot be even or odd
Even, because $f(-x)=f(x)$ (correct answer)
Neither, because it contains both even and odd powers of $x$
Odd, because $f(-x)=-f(x)$

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! To check, compute f(-x) = (-x)^4 - 2(-x)^2 = x^4 - 2x^2, which equals f(x), confirming y-axis symmetry. Choice A correctly determines it's even because f(-x) = f(x). If you chose C thinking mixed powers mean neither, that's understandable, but actually all powers here are even, making it even—odd powers would suggest possible oddness! For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither. Example: f(x) = x² - 3, so f(-x) = (-x)² - 3 = x² - 3 = f(x) → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.

Question 5

Given the graph of $f(x)=x^2$ and the graph of $g(x)=f(x)+k$ shown, the vertex of $g$ is at $(0,-3)$ . What is $k$ ?

$k=3$
$k=-\tfrac{1}{3}$
$k=\tfrac{1}{3}$
$k=-3$ (correct answer)

Explanation: This question tests your understanding of how algebraic transformations of functions—like

f(x) + k

k \cdot f(x)

f(kx)

, and

f(x + k)

—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1)

f(x) + k

shifts the graph vertically (up if

k > 0

, down if

k < 0

), (2)

k \cdot f(x)

stretches vertically if

|k| > 1

or compresses if

0 < |k| < 1

(and reflects across x-axis if

k < 0

), (3)

f(x + k)

shifts horizontally—LEFT if

k > 0

, RIGHT if

k < 0

(opposite of what you might expect!), (4)

f(kx)

compresses horizontally if

|k| > 1

or stretches if

0 < |k| < 1

(and reflects across y-axis if

k < 0

). Outside the function (

f(x) + k

and

k \cdot f(x)

) affects y-values/vertical; inside the function (

f(x + k)

and

f(kx)

) affects x-values/horizontal. Here,

g(x) = f(x) + k

shifts vertically by

k

, moving the vertex from

(0,0)

(0,k)

, so for

(0,-3)

k

must be -3 to shift down 3 units. Choice B correctly finds the

k

value as -3. If you selected A, you might have forgotten that positive

k

shifts up, but negative shifts down—keep that sign in mind! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like

f(x) + 3

2f(x)

) changes y-values (vertical effects). Anything done INSIDE the parentheses (like

f(x + 3)

f(2x)

) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—

f(x + 3)

shifts LEFT 3 because you're subtracting 3 from x-coordinates, and

f(x - 2)

shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 6

If $g(x) = f(kx)$ where $k > 0$ , and the period of $f(x)$ is 8, what value of $k$ would make the period of $g(x)$ equal to 2?

$k = \frac{1}{4}$
$k = \frac{1}{2}$
$k = 2$
$k = 4$ (correct answer)

Explanation: When

f(x)

has period

P

, then

f(kx)

has period

\frac{P}{k}

. Given that

f(x)

has period 8,

g(x) = f(kx)

has period

\frac{8}{k}

. We want this period to equal 2, so

\frac{8}{k} = 2

, which gives

k = 4

. Choice A gives period

\frac{8}{1/4} = 32

. Choice B gives period

\frac{8}{1/2} = 16

. Choice C gives period

\frac{8}{2} = 4

Question 7

Consider the function $f(x) = \sqrt{x}$ with domain $x \geq 0$ . What is the domain of $g(x) = f(2-x) + 3$ ?

$x \geq 2$
$x \leq 2$ (correct answer)
$x \geq -2$
$x \leq -2$

Explanation: For

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

to be defined, we need the expression under the square root to be non-negative:

2-x \geq 0

, which gives

x \leq 2

. The vertical shift

+3

doesn't affect the domain. Choice A reverses the inequality. Choice C would be correct if we had

f(x+2)

instead. Choice D combines both errors.

Question 8

If $f(x)$ is a function where $f(2) = 5$ , and $g(x) = 3f(\frac{x}{2}) - 1$ , what is the value of $g(4)$ ?

$11$
$14$ (correct answer)
$17$
$29$

Explanation: To find

g(4)

, substitute into

g(x) = 3f(\frac{x}{2}) - 1

g(4) = 3f(\frac{4}{2}) - 1 = 3f(2) - 1

. Since

f(2) = 5

, we have

g(4) = 3(5) - 1 = 15 - 1 = 14

. Choice A results from computing

2f(2) + 1

. Choice C results from computing

3f(2) + 2

. Choice D results from computing

6f(2) - 1

Question 9

The function $p(x)$ satisfies $p(-x) = -p(x)$ for all $x$ . If $h(x) = p(x-3) + 2$ , which property does $h(x)$ possess?

$h(x)$ is even: $h(-x) = h(x)$
$h(x)$ is odd: $h(-x) = -h(x)$
$h(x)$ is symmetric about the point $(3, 2)$ (correct answer)
$h(x)$ is symmetric about the line $x = 3$

Explanation: Since

p(-x) = -p(x)

p(x)

is odd and symmetric about the origin. For

h(x) = p(x-3) + 2

, this shifts the graph right 3 units and up 2 units. An odd function shifted right and up becomes symmetric about the point where the origin moves to, which is

(3, 2)

. To verify:

h(3+t) + h(3-t) = p(t) + 2 + p(-t) + 2 = p(t) - p(t) + 4 = 4 = 2 \cdot 2

, confirming point symmetry about

(3,2)

. Choice A is wrong because horizontal shifts destroy evenness. Choice B is wrong because vertical shifts destroy oddness. Choice D describes line symmetry, not point symmetry.

Question 10

The function $f(x) = x^2$ is transformed to create $g(x) = -2f(x-3) + 1$ . If the vertex of $f(x)$ is at $(0,0)$ , what are the coordinates of the vertex of $g(x)$ ?

$(3, 1)$ (correct answer)
$(-3, -1)$
$(3, -1)$
$(-3, 1)$

Explanation: The transformations applied are: horizontal shift right 3 units (from

x-3

), vertical stretch by factor 2 and reflection over x-axis (from

-2

), and vertical shift up 1 unit (from

+1

). Starting with vertex

(0,0)

: horizontal shift gives

(3,0)

, the vertical transformations affect only the y-coordinate giving

(3, -2(0)+1) = (3,1)

. Choice B reflects the horizontal shift incorrectly. Choice C ignores the vertical shift. Choice D makes errors in both horizontal direction and final y-coordinate.

Question 11

If $f(x)$ is an even function and $g(x) = f(2x-4)$ , which statement about $g(x)$ is true?

$g(x)$ is even and symmetric about the y-axis
$g(x)$ is odd and symmetric about the origin
$g(x)$ is neither even nor odd, but symmetric about $x=2$ (correct answer)
$g(x)$ is neither even nor odd, but symmetric about $x=4$

Explanation: Since

f(x)

is even,

f(-x) = f(x)

. For

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

, we can rewrite this as

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

. This represents a horizontal compression by factor

\frac{1}{2}

followed by a horizontal shift right by 2 units. The compression preserves evenness, but the horizontal shift destroys it. However, the function maintains symmetry about the line

x=2

. To verify:

g(2+h) = f(2(2+h)-4) = f(2h)

and

g(2-h) = f(2(2-h)-4) = f(-2h) = f(2h)

since

f

is even. Choice A is wrong because the shift destroys evenness about y-axis. Choice B is wrong because

g

is not odd. Choice D has the wrong axis of symmetry.

Question 12

Determine whether the function $p(x)=x^3-x$ is even, odd, or neither.

Neither even nor odd
Odd (correct answer)
Both even and odd
Even

Explanation: This question tests your understanding of how algebraic transformations of functions—like

f(x) + k

k \cdot f(x)

f(kx)

, and

f(x + k)

—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry:

f(-x) = f(x)

for all x, meaning the left half of the graph is a mirror image of the right half. Examples include

f(x) = x^2

x^4

|x|

, and

x^2 + 3

. Odd functions have origin symmetry:

f(-x) = -f(x)

, meaning rotating the graph 180° about the origin gives the same graph. Examples include

f(x) = x

x^3

1/x

, and

x^3 - x

. Most functions are neither even nor odd! For

p(x) = x^3 - x

, compute

p(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x)

, which equals

-p(x)

, confirming origin symmetry without matching

p(x)

. Choice B correctly determines the function is odd. An incorrect choice like A could stem from overlooking the signs after substitution—note the odd powers (3 and 1) flip signs appropriately for odd functions, but verification is key. For even/odd testing: (1) Take the given function

f(x)

, (2) Find

f(-x)

by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with

f(x)

and

-f(x)

: if

f(-x) = f(x)

, it's even; if

f(-x) = -f(x)

, it's odd; if neither match, it's neither. Example:

f(x) = x^2 - 3

, so

f(-x) = (-x)^2 - 3 = x^2 - 3 = f(x)

→ even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.

Question 13

Let $f(x)=\sqrt{x}$ (defined for $x\ge 0$ ). Consider $g(x)=f\left(\frac{x}{3}\right).$ How does the graph of $g$ compare to the graph of $f$ ?

Shift right 3 units
Vertical stretch by factor $3$
Horizontal compression by factor $3$
Horizontal stretch by factor $3$ (correct answer)

Explanation: This question tests your understanding of how algebraic transformations of functions—like

f(x) + k

k \cdot f(x)

f(kx)

, and

f(x + k)

—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1)

f(x) + k

shifts the graph vertically (up if

k > 0

, down if

k < 0

), (2)

k \cdot f(x)

stretches vertically if

|k| > 1

or compresses if

0 < |k| < 1

(and reflects across x-axis if

k < 0

), (3)

f(x + k)

shifts horizontally—LEFT if

k > 0

, RIGHT if

k < 0

(opposite of what you might expect!), (4)

f(kx)

compresses horizontally if

|k| > 1

or stretches if

0 < |k| < 1

(and reflects across y-axis if

k < 0

). Outside the function (

f(x) + k

and

k \cdot f(x)

) affects y-values/vertical; inside the function (

f(x + k)

and

f(kx)

) affects x-values/horizontal. For

g(x) = f(x/3)

with

f(x) = \sqrt{x}

, this is equivalent to

f((1/3)x)

, stretching the graph horizontally by a factor of 3 since

|1/3| < 1

, meaning x-values need to be three times larger to produce the same y (e.g.,

g(3) = f(1) = 1

, while

f(3) \approx 1.73

). Choice B correctly identifies this as a horizontal stretch by factor 3. A distractor like choice A confuses stretch with compression—when the coefficient inside is less than 1 in absolute value, it's a stretch, not compression; the opposite holds for

|k| > 1

. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like

f(x) + 3

2f(x)

) changes y-values (vertical effects). Anything done INSIDE the parentheses (like

f(x + 3)

f(2x)

) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—

f(x + 3)

shifts LEFT 3 because you're subtracting 3 from x-coordinates, and

f(x - 2)

shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 14

Let $f(x)=x^3-x$ . Show using symmetry rules whether $f$ is even, odd, or neither by considering $f(-x)$ relative to $f(x)$ .

Even, because $f(-x)=f(x)$
Odd, because $f(-x)=-f(x)$ (correct answer)
Neither, because $f(-x)\neq f(x)$ and $f(-x)\neq -f(x)$
Both even and odd, because $f(-x)=f(x)=-f(x)$

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! Computing f(-x) = (-x)^3 - (-x) = -x^3 + x = -(

x^3

- x) = -f(x), showing origin symmetry—amazing job with the algebra! Choice B correctly determines it's odd because f(-x) = -f(x). Choice A could mislead if you only check even powers, but the odd powers make f(-x) = -f(x), not f(x)—keep verifying both conditions! For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither. Example: f(x) = x² - 3, so f(-x) = (-x)² - 3 = x² - 3 = f(x) → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.

Question 15

The graph of $f(x)=x^2$ is shown. The graph of $g(x)$ has the same shape and vertex at $(0,-2)$ . Which equation matches $g(x)$ ?

$g(x)=2f(x)$
$g(x)=f(x-2)$
$g(x)=f(x)+2$
$g(x)=f(x)-2$ (correct answer)

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. Moving the vertex to (0,-2) with the same shape means subtracting 2 from the quadratic's y-values, a downward vertical shift—outstanding reasoning! Choice C correctly identifies g(x) = f(x) - 2. Choice A would shift up instead, to (0,2)—just flip the sign for vertical shifts, and you've nailed it! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 16

Let the parent function be $f(x)=\sqrt{x}$ . What is the effect on the graph of replacing $f(x)$ with $f\left(\frac{x}{2}\right)$ ?

Vertical stretch by a factor of 2
Shift right 2 units
Horizontal compression by a factor of 2
Horizontal stretch by a factor of 2 (correct answer)

Question 17

Let $f(x)=|x|$ . The graph of $g$ is shown along with $f$ . The function $g$ is of the form $g(x)=f(x)+k.$ What is the value of $k$ ?

(Use the graph: $f$ has vertex at $(0,0)$ and $g$ has vertex at $(0,3)$ .)

$k=-3$
$k=0$
$k=3$ (correct answer)
$k=\tfrac{1}{3}$

Explanation: This question tests your understanding of how algebraic transformations of functions—like

f(x) + k

k \cdot f(x)

f(kx)

, and

f(x + k)

—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1)

f(x) + k

shifts the graph vertically (up if

k > 0

, down if

k < 0

), (2)

k \cdot f(x)

stretches vertically if

|k| > 1

or compresses if

0 < |k| < 1

(and reflects across x-axis if

k < 0

), (3)

f(x + k)

shifts horizontally—LEFT if

k > 0

, RIGHT if

k < 0

(opposite of what you might expect!), (4)

f(kx)

compresses horizontally if

|k| > 1

or stretches if

0 < |k| < 1

(and reflects across y-axis if

k < 0

). Outside the function (

f(x) + k

and

k \cdot f(x)

) affects y-values/vertical; inside the function (

f(x + k)

and

f(kx)

) affects x-values/horizontal. Given

g(x) = f(x) + k

with

f(x) = |x|

and the graph showing g's vertex at (0,3) compared to f's at (0,0), this is a vertical shift up by 3 units, so

k = 3

directly since it's added outside the function. Choice C correctly finds the k value as 3. A mistake like choice A could come from confusing vertical with horizontal shifts or misreading the sign—since the vertex moved up, k is positive, not negative; down shifts would use negative k. Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like

f(x) + 3

2f(x)

) changes y-values (vertical effects). Anything done INSIDE the parentheses (like

f(x + 3)

f(2x)

) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—

f(x + 3)

shifts LEFT 3 because you're subtracting 3 from x-coordinates, and

f(x - 2)

shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 18

Show the symmetry type by identifying whether $f(x)=x^3-x$ is even, odd, or neither.

Even
Odd (correct answer)
Neither
Both even and odd

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. To test f(x) = x³ - x, we find f(-x) = (-x)³ - (-x) = -x³ + x = -(x³ - x) = -f(x), confirming it's odd. Choice B correctly identifies this function as odd (symmetric about the origin). The function cannot be even since f(-x) = -f(x) ≠ f(x), and it clearly satisfies the odd function definition. Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation. Most functions are neither even nor odd!

Question 19

Let $f(x)=\sqrt{x}$ be the parent function. Describe the effect of $g(x)=f(2x)$ on the graph of $f$ .

Horizontal stretch by a factor of 2
Vertical stretch by a factor of 2
Horizontal compression by a factor of 2 (correct answer)
Shift left 2 units

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Function transformations come in four main types: (1) f(x) + k shifts the graph vertically (up if k > 0, down if k < 0), (2) k·f(x) stretches vertically if |k| > 1 or compresses if 0 < |k| < 1 (and reflects across x-axis if k < 0), (3) f(x + k) shifts horizontally—LEFT if k > 0, RIGHT if k < 0 (opposite of what you might expect!), (4) f(kx) compresses horizontally if |k| > 1 or stretches if 0 < |k| < 1 (and reflects across y-axis if k < 0). Outside the function (f(x) + k and k·f(x)) affects y-values/vertical; inside the function (f(x + k) and f(kx)) affects x-values/horizontal. With g(x) = f(2x) for the square root function, multiplying x by 2 (inside) squeezes the graph horizontally toward the y-axis, making it steeper and narrower—excellent work recognizing this! Choice B correctly identifies the transformation as a horizontal compression by a factor of 2. Choice A might confuse compression with stretch, but since |2| > 1, it's compression, not stretch—you've got this! Transformation memory aid: think 'outside affects y, inside affects x.' Anything added/multiplied OUTSIDE f (like f(x) + 3 or 2f(x)) changes y-values (vertical effects). Anything done INSIDE the parentheses (like f(x + 3) or f(2x)) changes x-values (horizontal effects). The tricky part: horizontal shifts are opposite to the sign—f(x + 3) shifts LEFT 3 because you're subtracting 3 from x-coordinates, and f(x - 2) shifts RIGHT 2. Think: what x-value gives the original function's behavior?

Question 20

Determine whether the function $h(x)=x^4-2x^2$ is even, odd, or neither.

Odd
Even (correct answer)
Neither even nor odd
Both even and odd

Explanation: This question tests your understanding of how algebraic transformations of functions—like f(x) + k, k·f(x), f(kx), and f(x + k)—affect their graphs, and how to recognize even and odd functions from their symmetry properties. Even functions have y-axis symmetry: f(-x) = f(x) for all x, meaning the left half of the graph is a mirror image of the right half. Examples include f(x) = x², x⁴, |x|, and x² + 3. Odd functions have origin symmetry: f(-x) = -f(x), meaning rotating the graph 180° about the origin gives the same graph. Examples include f(x) = x, x³, 1/x, and x³ - x. Most functions are neither even nor odd! For h(x) = x⁴ - 2x², compute h(-x) = (-x)⁴ - 2(-x)² = x⁴ - 2x², which equals h(x), confirming y-axis symmetry without matching -h(x) = -x⁴ + 2x². Choice A correctly determines the function is even. A distractor like choice B might result from not fully simplifying h(-x) or confusing even with odd—note that the powers are all even (4 and 2), which often indicates even functions, but always verify algebraically. For even/odd testing: (1) Take the given function f(x), (2) Find f(-x) by substituting -x for every x (use parentheses!), (3) Simplify completely, (4) Compare with f(x) and -f(x): if f(-x) = f(x), it's even; if f(-x) = -f(x), it's odd; if neither match, it's neither. Example: f(x) = x² - 3, so f(-x) = (-x)² - 3 = x² - 3 = f(x) → even! Graphically: even functions have y-axis as mirror line, odd functions look the same after 180° rotation.