Exponential Functions & Graphs - Algebra 3

1. Let $g(x)=4-3^{2-x}$. Which statement about $g$ is true?
   Answer: $g$ is increasing, has range $(-\infty,4)$, and its x-intercept is $2-\log_3 4$. 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.
2. 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 $K$?
   Answer: $C=17{,}700, K=300$ 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.
3. The graph of $f(x)=4^x$ is horizontally stretched by a factor of 2. Which equation defines the transformed function $g$?
   Answer: $g(x)=4^{x/2}$ 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.
4. A student claims that because $3>2$, the graph of $g(x)=3^x$ is above the graph of $f(x)=2^x$ for every real $x$. Which statement best corrects the student's claim?
   Answer: The graphs intersect only at $x=0$; $g$ is below $f$ for $x<0$ and above $f$ for $x>0$. At $x=0$, $2^0=3^0=1$, so the graphs intersect. Since $g(x)/f(x)=(3/2)^x$, the ratio is less than 1 for $x<0$ and greater than 1 for $x>0$; because $f(x)>0$, $g$ is below $f$ for negative $x$ and above $f$ for positive $x$. The other choices ignore the negative interval, assert a false second intersection, or deny the intersection at $x=0$.
5. The function $f$ is defined by $f(x)=2^{x-3}+4$. Which of the following correctly describes its graph?
   Answer: 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 $\frac{33}{8}$. 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.
6. 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?
   Answer: Its horizontal asymptote is $y=1$, its y-intercept is 0, and its range is $y<1$. 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.
7. Let $f(x)=5(2)^{x+1}-10$. Which statement about the graph of $f$ is true?
   Answer: It is increasing, has range $(-10,\infty)$, and passes through the origin. Since $5(2)^{x+1}=10(2)^x$, $f$ is increasing. As $x\to-\infty$, $10(2)^x\to0$, so $f\to-10$, giving range $(-10,\infty)$. At $x=0$, $f(0)=10-10=0$, so the graph passes through the origin. The wrong choices reverse monotonicity, deny the x-intercept, or state the range incorrectly.
8. 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 $f(1)$?
   Answer: $-6$ 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 $12$.
9. 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 $f$?
   Answer: Reflect across the y-axis. 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 $-1$.
10. 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 $g$?
    Answer: $(0,1)$ corresponds to $(-2,-2)$ 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.
11. Let $f(x)=a(2)^x+b$, with $a<0$ and $b>0$. Which statement must be true about the graph of $f$?
    Answer: It is decreasing, has horizontal asymptote $y=b$, and has exactly one x-intercept. 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.
12. A single horizontal shift of $y=2^x$ produces the same graph as $g(x)=4\cdot 2^{x-3}$. Which shift is it?
    Answer: Shift right 1 unit. Using $4=2^2$, $g(x)=2^2\cdot2^{x-3}=2^{x-1}$. This is $2^x$ with $x$ replaced by $x-1$, so the same graph results from shifting right 1 unit. The other options come from ignoring the coefficient or using the wrong direction.
13. Let $f(x)=3^{x+1}-2$ and $g(x)=-f(x)+4$. Which statement about the graph of $g$ is true?
    Answer: $g$ is decreasing, has y-intercept $3$, horizontal asymptote $y=6$, and range $(-\infty,6)$. Since $f$ has horizontal asymptote $y=-2$, $g=-f+4$ has asymptote $y=6$. At $x=0$, $f(0)=1$, so $g(0)=3$. Because $f$ is increasing, $g$ is decreasing, and its range is $(-\infty,6)$. The choices that claim increasing, y-intercept $-3$, or asymptote $2$ misapply the reflection or vertical shift.
14. Which of the following functions is increasing for all real x and has a horizontal asymptote below the x-axis?
    Answer: $f(x)=2(3^x)-4$ 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.
15. Which function is decreasing for all $x$ and has no x-intercept?
    Answer: $f(x)=3\left(\frac12\right)^x+3$ $3(1/2)^x+3$ 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 $x=0$.
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?
    Answer: $p(x)=3-4^{2-x}$ Reflecting $4^x$ across the y-axis gives $4^{-x}$. Shifting right 2 replaces $x$ by $x-2$, 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.
17. 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?
    Answer: $g(x)=3^x+2$ 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 $f(x)=\left(\frac13\right)^x$ as $3^{-x}$, 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 $\frac13$ as $3^{-1}$ 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.
18. Which of the following functions has a graph identical to $h(x)=2^{x+2}$?
    Answer: $h(x)=4(2^x)$ 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 $a^{bx+c}$ or $k\cdot a^x$ 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.
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 $b$?
    Answer: $a=2, b=\frac12$ 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.
20. 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?
    Answer: $P(t)=500(2)^{t/3}$; 9 hours 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.

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