SAT nonlinear functions and exponentials: review and practice

• Read the vertex, zeros, intercepts and maximum or minimum from a function's equation or graph.
• Match a nonlinear equation to its graph or table.
• Write and interpret exponential growth and decay models, like 3% growth per year.
• Use function notation and shift graphs left, right, up or down.
• Explain what a zero, an intercept or a maximum means in a real situation.

• Giving the time when the question asks for the height. In h(t) = −5(t − 3)² + 50, the vertex is (3, 50): the maximum height is 50, reached at t = 3.
• Writing the growth rate instead of the growth factor. Growing 4% a year means multiplying by 1.04 each year, not by 0.04 or 4.
• Shifting the wrong way. f(x − 3) moves the graph 3 units right, not left, and f(x) − 3 moves it 3 units down.
• Confusing a zero with the y-intercept. Zeros are the x-values where f(x) = 0. The y-intercept is f(0).
• Using the multiplier for the wrong time period. If a population doubles every 6 hours and t is in minutes, the exponent is t/360, not t/6.
• Ignoring the domain. If a model only applies for 0 ≤ t ≤ 10, the maximum may be at an endpoint rather than the vertex.

1. Question 1 · Easy

   The function h is defined by h(t) = −4(t − 2)² + 21 and models the height, in meters, of a ball t seconds after it is thrown. Which of the following is the best interpretation of the vertex of the graph of y = h(t) in this context?

   • AThe ball reaches a minimum height of 21 meters 2 seconds after it is thrown.
   • BThe ball reaches a maximum height of 2 meters 21 seconds after it is thrown.
   • CThe ball reaches a minimum height of 2 meters 21 seconds after it is thrown.
   • DThe ball reaches a maximum height of 21 meters 2 seconds after it is thrown.
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   D. The negative coefficient makes the vertex a maximum, and the vertex (2, 21) means a height of 21 meters after 2 seconds. Choice A calls the maximum a minimum, choice B exchanges time and height, and choice C makes both errors.

2. Question 2 · Easy

   A concert venue sold 18,600 season passes this year. A model predicts that pass sales will increase by 2.8% each year relative to the previous year. Which equation represents the predicted number of passes, S(t), sold t years from this year?

   • AS(t) = 18600(0.28)ᵗ
   • BS(t) = 18600(1.28)ᵗ
   • CS(t) = 18600(1.028)ᵗ
   • DS(t) = 18600(0.028)ᵗ
   Show the answer

   C. An annual increase of 2.8% gives a multiplier of 1 + 0.028 = 1.028, so S(t) = 18600(1.028)ᵗ; this gives 18,600 at t = 0 and 19,120.8 at t = 1. Choice D omits the original amount from the multiplier, choice A also misplaces the decimal, and choice B models a 28% increase.

3. Question 3 · Medium

   The function r is defined by r(x) = 18/x² − 11. The graph of y = t(x) is obtained by translating the graph of y = r(x) a units to the right and b units down, where a and b are positive constants. Which equation defines t?

   • At(x) = 18/(x − a)² − (11 − b)
   • Bt(x) = 18/(x − a)² − (11 + b)
   • Ct(x) = 18/(x² − a) − (11 + b)
   • Dt(x) = 18/(x + a)² − (11 + b)
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   B. The translation gives t(x) = r(x − a) − b = 18/(x − a)² − 11 − b = 18/(x − a)² − (11 + b). The other expressions subtract a after squaring, reverse the horizontal direction, or change the downward shift into an upward shift.

4. Question 4 · Medium

   The function h is defined by h(x) = 2(x + p)(x + 3)(3x − 18), where p is a constant. In the xy-plane, y = h(x) passes through the point (−5, 0). What is the value of h(0)?

   • A−540
   • B−270
   • C5
   • D540
   Show the answer

   A. At x = −5, the last two factors are −2 and −33, so p − 5 = 0 and p = 5; therefore h(0) = 2(5)(3)(−18) = −540. Choice B omits the multiplier 2, choice C stops at the parameter, and choice D drops the negative sign of the final factor.

5. Question 5 · Hard

   An underwater glider’s depth below the surface, in meters, is modeled by a quadratic function of time after observations begin. Its greatest depth is 192 meters. At 8 minutes after observations begin, the glider is 144 meters below the surface and is moving toward the surface. The glider reaches the surface 12 minutes after observations begin. According to the model, what is the glider’s depth, in meters, 10 minutes after observations begin?

   • A72
   • B84
   • C108
   • D180
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   B. Write d(t) = 192 − a(t − h)², where a > 0 and h < 8; the observations give a(12 − h)² = 192 and a(8 − h)² = 48, so (8 − h)/(12 − h) = 1/2, h = 4, and a = 3. Hence d(10) = 192 − 3(10 − 4)² = 84; choice A linearly interpolates between the two observations, choice C reports the decrease from the maximum, and choice D uses the other algebraic branch, whose maximum occurs after minute 8.

6. Question 6 · Hard

   A population is modeled by N(t) = 28Q^(t/9), where t is measured in minutes and Q > 1. The model predicts a p% increase during every 180-second interval. What is the value of p, expressed using Q?

   • A100(Q²⁰ + 1)
   • B100(Q^(1/3) - 1)
   • C100(Q²⁰ - 1)
   • D100(Q^(1/3) + 1)
   Show the answer

   B. An interval of 180 seconds is 3 minutes, so its output ratio is Q^(3/9) = Q^(1/3) and the increase is p = 100(Q^(1/3) - 1). Choice D adds the original proportion instead of subtracting it, choice C leaves seconds unconverted, and choice A combines those errors.

Define the function first, like f(x) = −5(x − 3)² + 50, then type f(4) on the next line to evaluate it. Click the graph to see the vertex and intercepts as gray points with coordinates. For an exponential model with mixed units, put the conversion inside the exponent: for a model that doubles every 6 hours with x in minutes, the exponent is x/360.

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