AP Precalculus: Exponential Functions (Drill 1)

Question 1. The function \( f(x) = 3 \cdot (0.7)^x \) is defined for all real numbers. Which of the following statements correctly describes \( f \)?

• A) The function is increasing, and its range is all real numbers.
• B) The function is decreasing, and its range is \( y > 0 \). ✓
• C) The function is increasing, and its range is \( y > 0 \).
• D) The function is decreasing, and its range is \( y \geq 3 \).

Explanation: Choice B is correct. Since \( b = 0.7 \) satisfies \( 0 < b 0 \), all output values are positive but never reach zero. The range is \( y > 0 \) because the x-axis is a horizontal asymptote the function approaches but never touches.

Choice A is incorrect because \( b = 0.7 < 1 \) signals decay, not growth.

Choice C is incorrect for the same reason: the function is decreasing, not increasing.

Choice D is incorrect because the range is \( y > 0 \), not \( y \geq 3 \). The value \( y = 3 \) is the y-intercept, \( f(0) = 3 \), not a range boundary.

Question 2. A population is growing at a rate of 8% per year. Which of the following functions correctly models a population of 500 that grows at this rate for \( t \) years?

• A) \( P(t) = 500 \cdot (0.08)^t \)
• B) \( P(t) = 500 \cdot (0.92)^t \)
• C) \( P(t) = 500 \cdot (1.08)^t \) ✓
• D) \( P(t) = 500 \cdot (1.8)^t \)

Explanation: Choice C is correct. An 8% annual growth rate means the population is multiplied by \( 1 + 0.08 = 1.08 \) each year. Starting from 500, the model is \( P(t) = 500 \cdot (1.08)^t \).

Choice A is incorrect because \( b = 0.08 \) is the growth rate, not the growth factor.

Choice B is incorrect because \( b = 0.92 = 1 - 0.08 \) represents an 8% decay rate, not growth.

Choice D is incorrect because \( b = 1.8 \) would represent an 80% growth rate, not 8%.

Question 3.

x	f(x)
0	4
1	12
2	36
3	108

Which of the following functions is consistent with the table above?

• A) \( f(x) = 4x + 8 \)
• B) \( f(x) = 3 \cdot 4^x \)
• C) \( f(x) = 4 \cdot 3^x \) ✓
• D) \( f(x) = 4 \cdot 3^x + 8 \)

Explanation: Choice C is correct. Check that the ratio of consecutive outputs is constant: \( 12/4 = 3 \), \( 36/12 = 3 \), \( 108/36 = 3 \). The common ratio is \( b = 3 \). The initial value at \( x = 0 \) is 4, so \( a = 4 \). This gives \( f(x) = 4 \cdot 3^x \).

Choice A is incorrect because a linear function produces constant differences, not constant ratios. The differences here are 8, 24, and 72, not constant.

Choice B is incorrect because \( f(x) = 3 \cdot 4^x \) gives \( f(0) = 3 \neq 4 \) and \( f(2) = 48 \neq 36 \).

Choice D is incorrect because adding 8 shifts the function vertically, giving \( f(0) = 12 \neq 4 \).

Question 4. Let \( f(x) = 5 \cdot 2^x \). A student defines a new function \( g \) by keeping the same coefficient but replacing \( x \) with its opposite before evaluating the power. Which of the following correctly identifies the transformation and its geometric effect on the graph of \( f \)?

• A) \( g \) is a reflection of \( f \) across the x-axis, so \( g \) is always negative.
• B) \( g \) is a vertical stretch of \( f \) by a factor of 2, so \( g \) grows faster than \( f \).
• C) \( g \) is a horizontal shift of \( f \) to the right by 2 units, so the two graphs have the same shape but different positions.
• D) \( g \) is a reflection of \( f \) across the y-axis, so where \( f \) is increasing, \( g \) is decreasing. ✓

Explanation: Choice D is correct. Replacing \( x \) with \( -x \) gives \( g(x) = 5 \cdot 2^{-x} \), reflecting the graph across the y-axis. Since \( f \) is increasing, its y-axis reflection is decreasing.

Choice A is incorrect because a reflection across the x-axis negates the output, giving \( -5 \cdot 2^x \). That function is always negative, but \( g \) is always positive.

Choice B is incorrect because a vertical stretch by 2 gives \( 10 \cdot 2^x \), not \( 5 \cdot 2^{-x} \).

Choice C is incorrect because a horizontal shift right by 2 gives \( 5 \cdot 2^{x-2} \), which still increases. The function \( g \) is decreasing.

Question 5. The function \( h(x) = -4 \cdot \left(\tfrac{1}{2}\right)^x + 6 \) is an exponential function. Which of the following correctly states the range of \( h \)?

• A) \( y < 6 \) ✓
• B) \( y > 6 \)
• C) \( y > -4 \)
• D) All real numbers

Explanation: Choice A is correct. Begin with \( \left(\tfrac{1}{2}\right)^x \), range \( y > 0 \). Multiplying by \( -4 \) reflects it across the x-axis, giving range \( y < 0 \). Adding 6 shifts the range up by 6, giving \( y < 6 \). The horizontal asymptote moves from \( y = 0 \) to \( y = 6 \). Check: \( h(0) = -4 + 6 = 2 < 6 \) ✓.

Choice B is incorrect because the function approaches \( y = 6 \) from below, not above.

Choice C is incorrect because \( y > -4 \) mistakes the coefficient \( -4 \) for a vertical shift. The asymptote is at \( y = 6 \).

Choice D is incorrect because exponential functions always have a horizontal asymptote; their range is never all real numbers.