SAT Nonlinear functions

Quadratics and exponentials: vertices, intercepts, growth, decay.

12% of Math — the single most common Math typeMath · Advanced Math7 question types
~5per test

How to score it

  • Vertex form a(x − h)² + k hands you the vertex (h, k) directly.
  • “Decreases by 80%” → multiply by 0.2 each step; “increases by 8%” → ×1.08.
  • y-intercept: set x = 0 and evaluate.

Common traps

  • “Decreases by 80%” vs. “decreases to 80%” — the #1 exponential trap.
  • Reads the vertex with the wrong sign on h.
  • Confuses the growth factor with the percent itself.

The 7 question types, with real examples

Quadratic vertex / min / max

“What is the minimum value of [f(x)] / the x at which it occurs?”

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From the question bankMedium

The function f is defined by f(x) = -x2+4x+21x^{2} + 4x + 21.. What is the minimum value of f(x)?

  • A−25-25
  • B22
  • C2121
  • DThere is no minimum value.✓
Why D

Since the coefficient of x2x^{2} is negative, the parabola opens downward. A downward-opening parabola has a maximum value but no minimum value, as it extends to negative infinity.

Exponential growth / decay from context

“Which function models the relationship?”

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From the question bankMedium

A population of bacteria starts at 500 and doubles every 3 hours. Which function models the number of bacteria, b, after h hours?

  • Ab(h)=500(2)h3\displaystyle b(h) = 500(2)^{\frac{h}{3}}✓
  • Bb(h)=500(2)3hb(h) = 500(2)^{3h}
  • Cb(h)=500(3)h2\displaystyle b(h) = 500(3)^{\frac{h}{2}}
  • Db(h)=500+2h3\displaystyle b(h) = 500 + \frac{2h}{3}
Why A

The population doubles every 3 hours, so the growth factor is 2 applied every 3 hours. After h hours, the number of 3-hour periods is h/3h/3, giving b(h)=500b(h) = 500(2)^(h/3)(h/3).

Intercepts of a quadratic / exponential

“What is the y-intercept of the graph of y = f(x)?”

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From the question bankMedium

The function f is defined by f(x)=x2−6x+5f(x) = x^{2} - 6x + 5.. What is the y-intercept of the graph of y=fy = f(x) in the xy-plane?

  • A−5-5
  • B33
  • C55✓
  • D66
Why C

The y-intercept occurs when x=0x = 0. Substituting x=0x = 0 into f(x) = x2x^{2} - 6x+56x + 5 gives f(0)=02−6f(0) = 0^{2} - 6(0) + 5 = 5.

Function transformation / shift identification

“[FIGURE: graph of f]. Which of the following could be the equation of f?”

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From the question bankMedium

The function f is defined by f(x)=(x+2)2−9f(x) = (x + 2)^{2} - 9.. In the xy-plane, the graph of y=ky = k(x) is the result of shifting the graph of y=fy = f(x) up 6 units. Which equation defines function k?

  • Ak(x)=(x+8)2−9k(x) = (x + 8)^{2} - 9
  • Bk(x)=(x+2)2−3k(x) = (x + 2)^{2} - 3✓
  • Ck(x)=(x+2)2+6k(x) = (x + 2)^{2} + 6
  • Dk(x)=(x+2)2−15k(x) = (x + 2)^{2} - 15
Why B

Shifting a graph up 6 units means adding 6 to the function value. Starting with f(x)=(x+2)2−9f(x) = (x + 2)^{2} - 9, we get k(x)=f(x)+6=(x+2)2−9+6=(x+2)2−3k(x) = f (x) + 6 = (x + 2)^{2} - 9 + 6 = (x + 2)^{2} - 3.

Interpret a function value statement

“What is the best interpretation of the statement "[f(a) is approximately equal to b]" in this context?”

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From the question bankMedium

A chemist models the concentration of a compound in a solution using the function C, where C(h)=85(0.75)hC(h) = 85(0.75)^{h} gives the concentration in milligrams per liter h hours after the reaction begins.

What is the best interpretation of the statement "C(6) is approximately equal to 15" in this context?

  • A15 hours after the reaction begins, the concentration is approximately 6 milligrams per liter.
  • BThe concentration decreases by approximately 15 milligrams per liter every 6 hours.
  • C6 hours after the reaction begins, the concentration is approximately 15 milligrams per liter.✓
  • D6 hours after the reaction begins, the concentration is approximately 150 milligrams per liter.
Why C

The function C(h) gives the concentration in milligrams per liter after h hours. The statement C(6) ≈ 15 means when h=6h = 6, the concentration C is approximately 15. Therefore, 6 hours after the reaction begins, the concentration is approximately 15 milligrams per liter.

Determine parameter from given function values

“The function f is defined parametrically; the graph of y = f(x) passes through [(0, 10) and (-2, 325/36)]. What is the value of ab?”

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From the question bankMedium

The function f is defined by f(x)=4x2−16x+kf(x) = 4x^{2} - 16x + k, where k is a constant. If the minimum value of f(x) is 3, what is the value of k?

  • A33
  • B1616
  • C1919✓
  • D−13-13
Why C

The x-coordinate of the vertex is at x=−(−16)/(2⋅4)=16/8=2x = -(-16)/(2\cdot 4) = 16/8 = 2. The minimum value occurs at this x-coordinate, so f(2)=3f(2) = 3. Substituting: 4(2)2−16(2)+k=34(2)^{2} - 16 (2) + k = 3, which gives 16−32+k=316 - 32 + k = 3, so k=19k = 19.

Evaluate a nonlinear function at a given input

“The function f is defined by f(x) = [EXPRESSION]. What is the value of f([k])? (inverse variant: For what value of x does f(x) = [N]?)”

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From the question bankMedium

The function f is defined by f(x)=3(x+2)(x−5)f(x) = 3 (x + 2) (x - 5). What is the value of f(1)?

  • A−36-36✓
  • B−12-12
  • C1212
  • D3636
Why A

Substitutingx=1intof(x)=3(x+2)(x−5)givesf(1)=3(1+2)(1−5)=3(3)(−4)=−36.Substituting x = 1 into f(x) = 3(x + 2)(x - 5) gives f(1) = 3(1 + 2)(1 - 5) = 3(3)(-4) = -36.

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