Algebra 3 Quiz: Exponential Functions And Graphs

20 questions · exam conditions

0:00

Exponential Functions And GraphsQuestion 1 of 20

The function $f$ is defined by $f(x)=2^{x-3}+4$ . Which of the following correctly describes its graph?

It is the graph of

2^x

shifted 3 units left and 4 units up; its horizontal asymptote is

y=4

It is the graph of

2^x

shifted 3 units right and 4 units up; its horizontal asymptote is

y=4

, and its y-intercept is

\frac{33}{8}

It is the graph of

2^x

shifted 3 units right and 4 units down; its horizontal asymptote is

y=-4

, and its y-intercept is

-\frac{31}{8}

It is the graph of

2^x

shifted 3 units left and 4 units down; its horizontal asymptote is

y=-4

, and its y-intercept is

-\frac{31}{8}

← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Exponential Functions And Graphs

Practice Exponential Functions And Graphs 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 Exponential Functions And Graphs, 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

The function $f$ is defined by $f(x)=2^{x-3}+4$ . Which of the following correctly describes its graph?

It is the graph of $2^x$ shifted 3 units left and 4 units up; its horizontal asymptote is $y=4$ .
It is the graph of $2^x$ shifted 3 units right and 4 units up; its horizontal asymptote is $y=4$ , and its y-intercept is $\frac{33}{8}$ . (correct answer)
It is the graph of $2^x$ shifted 3 units right and 4 units down; its horizontal asymptote is $y=-4$ , and its y-intercept is $-\frac{31}{8}$ .
It is the graph of $2^x$ shifted 3 units left and 4 units down; its horizontal asymptote is $y=-4$ , and its y-intercept is $-\frac{31}{8}$ .

Explanation: Whenever you see an exponential written as

f(x)=2^{x-3}+4

, think of the general transformation form

a^{x-h}+k

. Here

h=3

, so replacing

x

with

x-3

shifts the graph 3 units to the right;

k=4

, so the graph shifts 4 units up, and the horizontal asymptote moves to

y=4

. Checking the

y

-intercept:

f(0)=2^{-3}+4=\frac18+4=\frac{33}{8}

, which confirms the correct description. The "shifted 3 units left and 4 units up" version misreads the exponent sign:

x-3

is right, not left. The "shifted 3 units right and 4 units down" version gets the horizontal shift right, but a

-4

outside would be needed for down; that would put the asymptote at

y=-4

and

y

-intercept at

-\frac{31}{8}

. The "shifted 3 units left and 4 units down" version has both wrong shift directions, though its asymptote and intercept do match a downward shift of

2^{x-3}

, not of

2^{x-3}+4

. So the reliable quick check is to identify

h

and

k

from

a^{x-h}+k

: right/up for positive

h,k

, and always plug in

x=0

to verify the

y

-intercept. That catches nearly every trap in these exponential transformation questions.

Question 2

The graph of $g$ is obtained from the graph of $f(x)=3^x$ by shifting it 2 units right and 1 unit up. Which of the following statements is true?

$g(x)=3^{x-2}+1$ , and the point $(2,2)$ is on the graph of $g$ . (correct answer)
$g(x)=3^{x+2}+1$ , and the point $(-2,2)$ is on the graph of $g$ .
$g(x)=3^{x-2}-1$ , and the point $(2,0)$ is on the graph of $g$ .
$g(x)=3^{x+2}-1$ , and the point $(-2,0)$ is on the graph of $g$ .

Explanation: Whenever you see a transformation of an exponential function, focus on two things: how the formula is changed by the shift, and whether the given point actually satisfies that formula. A horizontal shift happens inside the exponent: replacing

x

with

x-2

shifts the graph 2 units right. A vertical shift happens outside: adding 1 moves the graph up 1 unit. So the correct function is

g(x)=3^{x-2}+1

. To verify the point, substitute

x=2

g(2)=3^{0}+1=2

, so

(2,2)

is indeed on the graph. Now examine the wrong formulas.

g(x)=3^{x+2}+1

uses

x+2

, which shifts the graph 2 units left, not right.

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

has the correct horizontal shift but subtracts 1 outside, shifting down instead of up.

g(x)=3^{x+2}-1

combines both errors: left and down. Interestingly, the points given in those choices, such as

(-2,2)

(2,0)

, do satisfy their own formulas, so they are traps: a point being on the graph does not mean the transformation was applied correctly. Your study tip: when checking a shifted exponential function, first determine whether the exponent is

x-h

x+h

, then check whether the constant outside is added or subtracted. Finally, test the point by plugging it in — both the formula and the point must match the described shifts.

Question 3

Which of the following functions is increasing for all real x and has a horizontal asymptote below the x-axis?

$f(x)=-2(3^x)-4$
$f(x)=2(3^x)-4$ (correct answer)
$f(x)=2\left(\frac{1}{3}\right)^x-4$
$f(x)=-2(1/3)^x+4$

Explanation: Whenever you see an exponential function in the form

f(x)=a b^x+c

, focus on three features: the base

b

, the leading coefficient

a

, and the constant

c

. The horizontal asymptote is

y=c

, and the function is increasing when

a

and

\ln b

have the same sign. For

b>1

, that means

a>0

; for

0<b<1

, that means

a<0

. The function

f(x)=2(3^x)-4

has

b=3>1

and

a=2>0

, so it is increasing for all real

x

. Its constant term is

-4

, so the horizontal asymptote is

y=-4

, which is below the

x

-axis. Thus it satisfies both conditions. The choice

-2(3^x)-4

has the correct asymptote

y=-4

, but its coefficient is negative, so it is decreasing rather than increasing. Similarly,

2\left(\frac{1}{3}\right)^x-4

decreases because the base is less than 1 while the coefficient is positive. The function

-2(1/3)^x+4

is actually increasing, since a negative coefficient with a base less than 1 makes the derivative positive, but its asymptote is

y=4

, above the

x

-axis, so it fails the asymptote requirement. A quick study tip: check the constant term for the asymptote's location, then check the sign of the coefficient and whether the base is greater or less than 1 to determine increasing or decreasing behavior.

Question 4

The graph of $f(x)=5^x$ is shifted 2 units left and 3 units down to produce the graph of $g$ . Which point on the graph of $f$ corresponds to which point on the graph of $g$ ?

$(0,1)$ corresponds to $(-2,-2)$ (correct answer)
$(0,1)$ corresponds to $(2,-2)$
$(0,1)$ corresponds to $(-2,4)$
$(0,1)$ corresponds to $(2,4)$

Explanation: When you see a transformation question, think about what happens to a single point: shifting left or down means subtracting from the corresponding coordinate. The graph of

f(x)=5^x

contains the point

(0,1)

because

5^0=1

. A shift 2 units left moves the x-coordinate from 0 to

-2

. A shift 3 units down moves the y-coordinate from 1 to

-2

. So

(0,1)

corresponds to

(-2,-2)

. The choice

(2,-2)

correctly moves down 3 but moves right instead of left, treating "left" as adding to x. The choice

(-2,4)

correctly moves left 2 but moves up instead of down, treating "down" as adding to y. The choice

(2,4)

reverses both directions, as if the shift were right 2 and up 3. Each wrong option comes from mixing up the signs of the coordinate shifts. A useful habit: for a horizontal shift, "left" means subtract from x-coordinates, even though the function rule uses

x+2

inside. For a vertical shift, "down" means subtract from y-coordinates. Always test with a known point like the y-intercept

(0,1)

, and translate its coordinates directly by the shift. That will keep you from confusing the equation form with the actual movement of the graph.

Question 5

Which of the following functions has a graph identical to $h(x)=2^{x+2}$ ?

$h(x)=2^x+2$
$h(x)=2^{2x}$
$h(x)=2(4^x)$
$h(x)=4(2^x)$ (correct answer)

Explanation: Whenever you see an exponential expression like

2^{x+2}

, your first move should be to rewrite it using exponent rules. The key identity is

a^{m+n}=a^m\cdot a^n

. Here, that means

2^{x+2}=2^x\cdot 2^2=4(2^x)

. So the function with the identical graph is the one written as

4(2^x)

: it is just a vertical stretch of the parent exponential by a factor of 4, which produces the same curve as shifting

2^x

left by 2 units. Now look at the traps. The choice

2^x+2

adds 2 to the output, producing a vertical shift upward, not a horizontal shift left. The choice

2^{2x}

uses

2^{2x}=(2^x)^2=4^x

, which changes the base's growth rate entirely, so its graph is different. Similarly,

2(4^x)

is really

2(2^{2x})=2^{2x+1}

; it has a different exponent coefficient and is not equivalent to

2^{x+2}

. A reliable study tip: whenever you compare exponential functions, rewrite everything into a common form like

a^{bx+c}

k\cdot a^x

before deciding. Watch especially for the difference between "add to the exponent" and "add to the value"—multiplication by a constant in the exponent becomes multiplication by a power, never addition outside the exponent.

Question 6

Let $f(x)=-3(2^x)+4$ . Which statement about the graph of $f$ is true?

The range is $y>4$ , the y-intercept is 1, and $f$ is decreasing on its entire domain.
The range is $y<4$ , the y-intercept is 4, and $f$ is increasing on its entire domain.
The range is $y<4$ , the y-intercept is 1, and $f$ is decreasing on its entire domain. (correct answer)
The range is $y>4$ , the y-intercept is $-3$ , and $f$ is increasing on its entire domain.

Explanation: Whenever you see a transformed exponential like

f(x)=-3(2^x)+4

, focus on three features: the

y

-intercept, the direction of the graph, and the horizontal asymptote that determines the range. Evaluating at

x=0

gives

f(0)=-3(2^0)+4=-3+4=1

, so the

y

-intercept is 1, not 4 or

-3

. The coefficient

-3

is negative, so as

x

increases,

2^x

increases but is multiplied by a negative number, making the whole graph decrease on its entire domain. For the range, notice that

2^x>0

for all real

x

, so

-3(2^x)<0

and therefore

f(x)<4

. Also, as

x\to -\infty

2^x\to 0

, so the graph approaches 4 but never reaches it; as

x\to\infty

, the function drops without bound. Thus the true statement is the one saying the range is

y<4

, the

y

-intercept is 1, and the function is decreasing. The distractor with range

y>4

and decreasing flips the range across the asymptote. The one with range

y<4

y

-intercept 4, and increasing confuses the horizontal asymptote with the intercept and misses the effect of the negative coefficient. The last distractor combines range

y>4

, intercept

-3

, and increasing, mistaking the leading coefficient for the intercept and reversing both range and direction. For future problems, always test

f(0)

for the intercept and compare values like

f(0)

and

f(1)

to decide increasing versus decreasing. Then use the sign of the coefficient and the vertical shift to pin down the range.

Question 7

The graph of $f(x)=4^x$ is horizontally stretched by a factor of 2. Which equation defines the transformed function $g$ ?

$g(x)=4^{2x}$
$g(x)=2(4^x)$
$g(x)=4^{x/2}$ (correct answer)
$g(x)=4^{x-2}$

Explanation: When you see a transformation like a horizontal stretch, remember that it affects the input variable

x

inside the function, not the output. A horizontal stretch by a factor of 2 means every point moves twice as far from the

y

-axis, so the same

y

-value occurs at a larger

x

. To achieve that, you replace

x

with

\frac{x}{2}

in the original function. Thus,

f(x)=4^x

becomes

g(x)=4^{x/2}

. Check with a point: originally

f(1)=4

. After stretching, that

y

-value should appear at

x=2

, and indeed

g(2)=4^{2/2}=4^1=4

. So the correct choice is

g(x)=4^{x/2}

. Now look at the traps. The choice

4^{2x}

is a horizontal compression by a factor of

\frac{1}{2}

—it makes the graph narrower, not wider. The choice

2(4^x)

is a vertical stretch by 2, which multiplies the output, not the input. The choice

4^{x-2}

represents a horizontal shift to the right by 2 units, moving the graph without stretching it. Each of these confuses the direction or type of transformation. A solid strategy: for any horizontal transformation, always think about what happens to a single point. If the graph is stretched by factor

a

, then

x

becomes

\frac{x}{a}

. And remember, horizontal changes are the opposite of what intuition suggests—larger factor means dividing, not multiplying.

Question 8

A function of the form $f(x)=a b^x+c$ has horizontal asymptote $y=-2$ , y-intercept 1, and is increasing for all real $x$ . Which equation could define $f$ ?

$f(x)=-3(2^x)-2$
$f(x)=3\left(\frac{1}{2}\right)^x-2$
$f(x)=-3\left(\frac{1}{2}\right)^x+2$
$f(x)=3(2^x)-2$ (correct answer)

Explanation: Whenever you see an exponential of the form

f(x)=a b^x+c

, think about what each parameter controls:

c

is the horizontal asymptote,

a+c

is the y-intercept because

b^0=1

, and whether the function increases or decreases depends on both the sign of

a

and the size of

b

. Here, the horizontal asymptote

y=-2

forces

c=-2

. Since the y-intercept is 1, plug in

x=0

a(-2)=1

? Actually,

a\cdot 1+c=1

, so

a-2=1

, giving

a=3

. Finally, with

a=3

positive, the function is increasing only if

b>1

. So the matching equation is

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

. The choice

-3(2^x)-2

has the correct asymptote but a y-intercept of

-5

, and its negative coefficient makes it decreasing. The choice

3\left(\frac12\right)^x-2

has the right asymptote and y-intercept, but since

0<b<1

, it decreases. The choice

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

has asymptote

y=2

and y-intercept

-1

, so it misses both given features. Only

3(2^x)-2

satisfies all three conditions. Study tip: translate every feature into a parameter condition. Asymptote gives

c

, y-intercept gives

a+c

, and increasing/decreasing links the sign of

a

with whether

b>1

0<b<1

Question 9

A population of bacteria doubles every 3 hours. If the initial population is 500, which function models the population $P(t)$ after $t$ hours, and after how many hours does the population reach 4000?

$P(t)=500(2)^{3t}$ ; 1 hour
$P(t)=500(2)^{t/3}$ ; 3 hours
$P(t)=500(2)^{3t}$ ; 3 hours
$P(t)=500(2)^{t/3}$ ; 9 hours (correct answer)

Explanation: Whenever you see exponential growth like "doubles every hours," first identify the doubling period and express time as the number of periods. Here the bacteria double every 3 hours, so after

t

hours there are

t/3

doubling periods. Each period multiplies the population by 2, so the model is

P(t)=500(2)^{t/3}

. To reach 4000, solve

4000=500(2)^{t/3}

. Dividing gives

8=2^{t/3}

. Since

8=2^3

, you need

t/3=3

, so

t=9

hours. Now the wrong choices. The function

500(2)^{3t}

would mean

3t

doublings after

t

hours—that is, doubling every

1/3

hour, not every 3 hours. If you solve it, you get

t=1

hour, but that answer depends on the incorrect function. The choice

500(2)^{t/3}

with "3 hours" has the correct model but stops too soon: at 3 hours the population is

500\cdot 2=1000

, not 4000. It actually takes three 3-hour periods, or 9 hours, to double three times. The remaining option,

500(2)^{3t}

with "3 hours," combines both errors: it grows far too fast, and at 3 hours it would give

500\cdot2^9=256{,}000

, not 4000. A good habit is to check the exponent: for a doubling every

k

hours, use

t/k

, not

kt

. Then test your function at one known time, such as

P(3)=1000

, before solving for a target population.

Question 10

The graph of $f(x)=\left(\frac{1}{3}\right)^x$ is reflected across the y-axis and then shifted up 2 units. Which equation defines the resulting graph?

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

Explanation: When you see a transformation question involving exponential functions, ask yourself two things: what does a reflection across the

y

-axis do to the input, and does a vertical shift add outside the function? Start by rewriting

f(x)=\left(\frac13\right)^x

3^{-x}

, because

\frac13=3^{-1}

. Reflecting across the

y

-axis replaces every

x

with

-x

, so

f(-x)=3^{-(-x)}=3^x

. Then shifting up 2 units means adding 2 to the entire output, giving

g(x)=3^x+2

. That is the correct resulting graph. The choice

3^{-x}+2

shows the original graph shifted up but never reflected — it is still decreasing. The choice

3^x-2

has the correct reflection but shifts down instead of up, moving the graph in the wrong vertical direction. The choice

3^{-x}-2

combines both errors: it keeps the original decreasing shape and shifts down. Each wrong answer represents a common slip: forgetting that

(1/3)^x

already contains a negative exponent, confusing which sign means reflect across the

y

-axis, or mixing up the direction of a vertical shift. A strong strategy is to rewrite exponential bases like

\frac13

3^{-1}

before transforming. Then apply reflections to the exponent and vertical shifts to the whole expression. Finally, test a point: for

x=0

, the original

f(0)=1

, and after reflection and shifting up,

g(0)=1+2=3

, matching

3^x+2

. This quick check catches most transformation mistakes.

Question 11

The graph of $f(x)=a b^x$ , where $a\neq 0$ , $b>0$ , and $b\neq 1$ , passes through $(0,-2)$ and $(2,-18)$ . What is $f(1)$ ?

$-8$
$-6$ (correct answer)
$6$
$12$

Explanation: When you see an exponential function like

f(x)=a b^x

and two points, think: the point with

x=0

gives you the coefficient directly, because

b^0=1

. Here,

(0,-2)

tells you

a=-2

. Then plug the other point into

f(2)

-2 b^2 = -18

, so

b^2=9

, and since

b>0

and

b\neq 1

, you get

b=3

. Now the function is

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

, so

f(1) = -2 \cdot 3 = -6

. Why are the other choices traps?

-8

is the slope of the line through the two given points—a classic mistake if you treat the curve as linear instead of exponential.

6

comes from taking the geometric mean of

-2

and

-18

(which gives

6

) but forgetting the negative sign—the true value is

-6

12

appears if you mistakenly square the coefficient: using

a^2 b = 4 \cdot 3 = 12

instead of

a b^1

. Key strategy: for any exponential function, the y-intercept (where

x=0

) always reveals

a

, and a second point solves for the base

b

. Then evaluate at the requested

x

. Also, always check the sign of

a

—it tells you whether the outputs are positive or negative. Here, since

a

is negative,

f(1)

must be negative, so you can immediately rule out

6

and

12

Question 12

The graph of $f(x)=4^x$ is reflected across the x-axis and then translated 1 unit up. Which statement about the resulting graph is true?

Its horizontal asymptote is $y=-1$ , its y-intercept is 0, and its range is $y<-1$ .
Its horizontal asymptote is $y=1$ , its y-intercept is 1, and its range is $y>1$ .
Its horizontal asymptote is $y=-1$ , its y-intercept is 1, and its range is $y<-1$ .
Its horizontal asymptote is $y=1$ , its y-intercept is 0, and its range is $y<1$ . (correct answer)

Explanation: Whenever you see an exponential function being transformed, track three features step-by-step: the horizontal asymptote, the y-intercept, and the range. Start with

f(x)=4^x

, which has asymptote

y=0

, y-intercept

1

, and range

y>0

. Reflecting across the x-axis gives

y=-4^x

: the asymptote stays

y=0

, the y-intercept becomes

-1

, and the range flips to

y<0

. Translating up 1 unit gives

g(x)=-4^x+1

. So the asymptote moves to

y=1

, the y-intercept becomes

-1+1=0

, and since

-4^x

is always negative,

g(x)

is always less than

1

. Thus the range is

y<1

. The choice with asymptote

y=1

, y-intercept

0

, and range

y<1

is correct. The choice saying asymptote

y=-1

and range

y<-1

treats the translation as a shift down instead of up, or confuses which reflection was applied. The choice with y-intercept

1

and range

y>1

likely reflects across the y-axis instead of the x-axis, or forgets the reflection entirely. The remaining choice with asymptote

y=-1

and y-intercept

1

combines those errors, keeping the original y-intercept while misplacing the asymptote. A reliable study tip: apply transformations in order, and check one anchor point plus the asymptote. If the graph is reflected across the x-axis, the range flips below the asymptote; a vertical translation then shifts both the asymptote and the range boundary.

Question 13

Let $g(x)=4-3^{2-x}$ . Which statement about $g$ is true?

$g$ is increasing, has range $(4,\infty)$ , and has no x-intercept.
$g$ is decreasing, has range $(-\infty,4)$ , and its x-intercept is $2+\log_3 4$ .
$g$ is increasing, has range $(-\infty,4)$ , and has no x-intercept.
$g$ is increasing, has range $(-\infty,4)$ , and its x-intercept is $2-\log_3 4$ . (correct answer)

Explanation: Rewrite

g(x)=4-9(1/3)^x

. As

x

increases,

(1/3)^x

decreases, so

g

increases. As

x\to\infty

g\to4

, so the range is

(-\infty,4)

. Setting

g(x)=0

gives

3^{2-x}=4

, so

2-x=\log_3 4

, hence

x=2-\log_3 4

. The other choices reverse the sign in the log equation or falsely claim there is no x-intercept.

Question 14

For $b>1$ , which single transformation of the graph of $f(x)=b^x$ produces a decreasing exponential function whose y-intercept is the same as that of $f$ ?

Shift down by 1 unit.
Shift right by 1 unit.
Reflect across the x-axis.
Reflect across the y-axis. (correct answer)

Explanation: Reflecting

b^x

across the y-axis gives

b^{-x}=(1/b)^x

, which is decreasing and has y-intercept

1

, the same as

b^x

. A horizontal shift changes the y-intercept, and reflecting across the x-axis gives y-intercept

-1

Question 15

Which function is decreasing for all $x$ and has no x-intercept?

$f(x)=3\left(\frac12\right)^x+3$ (correct answer)
$f(x)=3\left(\frac12\right)^x-3$
$f(x)=3(2)^x+3$
$f(x)=3(2)^x-3$

Explanation:

3(1/2)^x+3

is decreasing and has range

(3,\infty)

, so it never reaches 0.

3(1/2)^x-3

crosses the x-axis at

x=0

;

3(2)^x+3

is increasing; and

3(2)^x-3

crosses the x-axis at

x=0

Question 16

Starting with $y=4^x$ , apply these transformations in order: reflect across the y-axis, shift right 2 units, reflect across the x-axis, shift up 3 units. Which equation defines the resulting function?

$p(x)=3-4^{x-2}$
$p(x)=3-4^{-x-2}$
$p(x)=3-4^{2-x}$ (correct answer)
$p(x)=3-4^{x+2}$

Explanation: Reflecting

4^x

across the y-axis gives

4^{-x}

. Shifting right 2 replaces

x

x-2

, giving

4^{-(x-2)}=4^{2-x}

. Reflecting across the x-axis and shifting up 3 gives

3-4^{2-x}

. The other formulas omit the y-axis reflection, shift in the wrong direction, or both.

Question 17

Which statement about $f(x)=2^{-x}-3$ is FALSE?

The graph is decreasing.
The graph has horizontal asymptote $y=-3$ .
The graph has x-intercept $-3$ . (correct answer)
The graph has y-intercept $-2$ .

Explanation: The graph is decreasing, has horizontal asymptote

y=-3

, and has y-intercept

-2

. Setting

2^{-x}-3=0

gives

2^{-x}=3

, so

-x=\log_2 3

, or

x=-\log_2 3

, not

-3

. Therefore the false statement is that the x-intercept is

-3

Question 18

Let $f(x)=a(2)^x+b$ , with $a<0$ and $b>0$ . Which statement must be true about the graph of $f$ ?

It is decreasing, has horizontal asymptote $y=0$ , and has no x-intercept.
It is decreasing, has horizontal asymptote $y=b$ , and has exactly one x-intercept. (correct answer)
It is increasing, has horizontal asymptote $y=b$ , and has exactly one x-intercept.
It is decreasing, has horizontal asymptote $y=b$ , and has no x-intercept.

Explanation: Because

a<0

, as

x

increases the term

a(2)^x

decreases, so

f

is decreasing. As

x\to-\infty

2^x\to0

, so the horizontal asymptote is

y=b

. Since

b>0

, the range is

(-\infty,b)

, which contains 0, so continuity gives exactly one x-intercept. The wrong choices misstate the asymptote, monotonicity, or existence of an intercept.

Question 19

A function of the form $f(x)=a b^x+k$ has horizontal asymptote $y=3$ . Its graph contains $(0,5)$ and $(2,3.5)$ , and $f$ is decreasing. What are $a$ and $b$ ?

$a=2, b=\frac14$
$a=2, b=\frac12$ (correct answer)
$a=5, b=\frac12$
$a=2, b=2$

Explanation: The asymptote

y=3

gives

k=3

. From

(0,5)

a+3=5

, so

a=2

. From

(2,3.5)

2b^2+3=3.5

, so

b^2=0.25

. Since the function is decreasing,

b=\frac12

. Choosing

b=\frac14

treats

b^2

as if it were

b

; choosing

a=5

uses the y-intercept as the coefficient;

b=2

would make the function increasing.

Question 20

The value of a machine, in dollars, is modeled by $V(t)=C(0.9)^t+K$ , where $t$ is years since purchase. The value eventually approaches ($300), the initial value is $\$ 18{,}000 $, and after one year the value is$ $16{,}230 $. What are$ C $and$ K$?

$C=18{,}000, K=300$
$C=16{,}230, K=1{,}770$
$C=17{,}700, K=300$ (correct answer)
$C=17{,}700, K=0$

Explanation: As

t\to\infty

(0.9)^t\to0

, so the limiting value ($300) gives

K=300

. The initial value gives

C+300=18000

, so

C=17700

. Checking

t=1

0.9(17700)+300=16230

. Using

C=18000

treats the initial value as the coefficient;

K=1770

uses the first-year drop as the asymptote;

K=0

ignores the limiting value.