SAT Math Quiz: Solving Nonlinear Functions

20 questions · exam conditions

0:00

Solving Nonlinear FunctionsQuestion 1 of 20

Let $h(x)=|2x-5|$ . What is the value of $h(1)$ ?

-7

7

-3

3

← Back to quizzes

SAT Math Quiz

SAT Math Quiz: Solving Nonlinear Functions

Practice Solving Nonlinear Functions in SAT Math 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 Solving Nonlinear Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

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

Let $h(x)=|2x-5|$ . What is the value of $h(1)$ ?

$-7$
$7$
$-3$
$3$ (correct answer)

Explanation: The question asks for h(1) when h(x) = |2x - 5|. Substitute x = 1: h(1) = |2(1) - 5| = |2 - 5| = |-3| = 3, so the value 3 is correct, since absolute value reports distance from zero and is never negative. The value -3 is what you get by stopping one step early, evaluating 2x - 5 inside the bars and forgetting to apply the absolute value. The value 7 comes from adding instead of subtracting, treating the expression as |2 + 5|. And -7 combines both mistakes at once, adding rather than subtracting and then dropping the sign-clearing effect of the absolute value. Keep the order of operations straight: finish the arithmetic inside the bars first, then take the absolute value, which turns any negative result positive.

Question 2

On the coordinate plane, the curve shown crosses the x-axis at $x=-2$ and $x=3$ , and it is undefined at $x=1$ (a break occurs there). Which expression could represent a rational function with these features?

$\dfrac{(x+2)(x-3)}{x+1}$
$(x+2)(x-3)$
$\dfrac{(x+2)(x-3)}{x-1}$ (correct answer)
$\dfrac{x-1}{(x+2)(x-3)}$

Explanation: You need a rational function whose graph crosses the x-axis at x = -2 and x = 3 and is undefined at x = 1. A quotient equals zero where its numerator is zero, so (x+2) and (x-3) belong on top; it is undefined where its denominator is zero, so (x-1) belongs on the bottom. That is why (x+2)(x-3)/(x-1) works. The version with x+1 in the denominator has the right x-intercepts but breaks at x = -1, not x = 1, since x+1 = 0 when x = -1. The plain product (x+2)(x-3) is a polynomial defined for every value of x, so it has no break at all. The version with x-1 on top and (x+2)(x-3) on the bottom reverses everything: it crosses the axis at x = 1 and is undefined at x = -2 and x = 3.

Question 3

A car's value is modeled by $V(t)=18000(0.86)^t$ , where $t$ is the number of years since purchase. Which option best describes the rate of change per year?

Increases $14\%$ per year
Decreases $86\%$ per year
Decreases $14\%$ per year (correct answer)
Decreases $0.86\%$ per year

Explanation: This question asks about the rate of change for V(t) = 18000(0.86)^t. This is an exponential decay function because the base 0.86 is between 0 and 1. The base 0.86 means the value retains 86% of its previous value each year, which is equivalent to losing 14% each year (100% - 86% = 14%). The car's value decreases by 14% per year. A common error is confusing the decay factor (0.86) with the decay rate (0.14 or 14%). Another error is misinterpreting 0.86 as 0.86%. When analyzing exponential decay, remember that a base of (1 - r) corresponds to a decay rate of r.

Question 4

A function is defined by $k(x)=|2x-7|$ . What is the value of $k(1)$ ?

$-5$
$5$ (correct answer)
$9$
$3$

Explanation: This question asks to evaluate an absolute value function. The function k(x) = |2x - 7| needs to be evaluated at x = 1. Substituting: k(1) = |2(1) - 7| = |2 - 7| = |-5| = 5. The absolute value converts the negative result to positive. Common errors include forgetting to apply the absolute value or making arithmetic mistakes within the absolute value bars. Remember that absolute value always returns a non-negative result, so |-5| = 5.

Question 5

For $h(x)=\dfrac{x+4}{x^2-9}$ , which value of $x$ is NOT in the domain of $h$ ?

$-4$
$-3$ (correct answer)
$0$
$3$

Explanation: To find values not in the domain of h(x) = (x + 4)/(x² - 9), we need to identify where the denominator equals zero. Setting x² - 9 = 0, we get x² = 9, so x = ±3. Both x = 3 and x = -3 make the denominator zero and are therefore not in the domain. Among the given choices, x = -3 (option B) is the value that's not in the domain. The numerator x + 4 doesn't affect the domain since it can equal zero without causing issues. When finding domains of rational functions, always factor the denominator completely to find all values that make it zero.

Question 6

Consider the function $r(x)=\sqrt{x+5}$ . What is $r(4)$ ?

$3$ (correct answer)
$\sqrt{5}$
$\sqrt{1}$
$9$

Explanation: To find r(4) where r(x) =

\sqrt{x+5}

, substitute x = 4: r(4) =

\sqrt{4+5}

\sqrt{9}

= 3. A common error is mixing up the formula and computing

\sqrt{9-4}=\sqrt{5}

instead, or otherwise leaving the answer in radical form when it can be simplified. When evaluating square root functions, first compute the expression under the radical, then take the square root of the result.

Question 7

A function is defined by $m(x)=\sqrt{9-x}$ . What is the value of $m(5)$ ?

$2$ (correct answer)
$\sqrt{14}$
$4$
$\sqrt{5}$

Explanation: This question asks to evaluate a square root function. The function m(x) =

\sqrt{9-x}

needs to be evaluated at x = 5. Substituting: m(5) =

\sqrt{9-5}

\sqrt{4}

= 2. Common errors include leaving the answer in radical form when it can be simplified or making an arithmetic slip under the radical. Always simplify radicals to their simplest form when possible.

Question 8

A coordinate plane shows the graph of a function passing through the points $(-2,4)$ , $(0,0)$ , and $(2,4)$ , with a sharp corner at $(0,0)$ and symmetry about the $y$ -axis. Which equation best matches the graph?

$y=x^2$
$y=2x$
$y=|2x|$ (correct answer)
$y=\sqrt{x}$

Explanation: We need to match the described graph to an equation, noting the sharp corner at (0,0) and symmetry about the y-axis. The function passes through (-2,4), (0,0), and (2,4), showing that when |x| = 2, y = 4. Testing y = |2x|: when x = -2, y = |2(-2)| = |-4| = 4; when x = 0, y = |0| = 0; when x = 2, y = |2(2)| = 4. The sharp corner at the origin and the V-shape with symmetry about the y-axis are characteristic features of absolute value functions. Students often confuse this with a parabola, but parabolas have smooth curves, not sharp corners.

Question 9

On the coordinate plane, a curve passes through $(0,3)$ , decreases as $x$ increases, and approaches (but never reaches) the horizontal line $y=0$ . It also passes through $(1,1.5)$ . Which equation could represent the curve?

$y=3(0.5)^x$ (correct answer)
$y=3(1.5)^x$
$y=3-0.5x$
$y=\dfrac{3}{x}$

Explanation: The question requires selecting an equation for a curve passing through (0,3) and (1,1.5), decreasing toward y=0 without reaching it. This is an exponential decay function, distinguished by its asymptotic approach to the x-axis. Plug in x=0 to confirm y=3 matches the initial value, then x=1 gives y=3*(0.5)^1=1.5, fitting the point and decay behavior. The curve approaches y=0 as x increases, consistent with base between 0 and 1. Errors include choosing growth like (1.5)^x, which increases, or linear like 3-0.5x, which crosses y=0. Rational like 3/x passes (1,3) not (1,1.5). For such problems, test points in choices to verify fit and asymptotic behavior.

Question 10

The graph of $y=2^{x}-4$ is shown on a coordinate plane. Which is the range of this function?

$(-\infty,4)$
$(-4,\infty)$ (correct answer)
$(0,\infty)$
$[-4,\infty)$

Explanation: The question asks for the range of y = 2^x - 4. This is exponential, with 2^x > 0 for all real x, so y > -4. As x → ∞, y → ∞; as x → -∞, y → -4 but never reaches it, giving range (-4, ∞). It doesn't include -4 since 2^x ≠ 0. Errors include [ -4, ∞) by including asymptote, or (0,∞) ignoring shift. Linear would have all reals. Analyze base and transformations to find range for exponentials.

Question 11

A phone battery's charge is modeled by $C(t)=100(0.92)^t$ , where $t$ is hours since unplugged. Approximately what percent of the charge remains after $5$ hours?

$66\%$ (correct answer)
$92\%$
$37\%$
$100\%$

Explanation: This problem asks for the remaining battery charge after 5 hours using C(t) = 100(0.92)^t. This is an exponential decay function where 0.92 represents retaining 92% of the charge each hour (losing 8% per hour). To find the charge after 5 hours, we calculate C(5) = 100(0.92)^5 = 100(0.6591) ≈ 65.91%. The battery retains approximately 66% of its original charge. A common error is thinking 0.92^5 means losing 8% × 5 = 40%, leaving 60%, but exponential decay compounds the loss each hour. For quick estimation on tests, remember that (0.92)^5 is slightly less than (0.9)^5 ≈ 0.59, so 66% is reasonable.

Question 12

A square-root function is defined by $f(x)=\sqrt{5-2x}$ . What is the domain of $f$ in real numbers? Choose the option that correctly applies the restriction on the expression under the radical.

$x\le \tfrac{5}{2}$ (correct answer)
$x<\tfrac{5}{2}$
$x\ge \tfrac{5}{2}$
$(-\infty,\infty)$

Explanation: This question asks for the domain of f(x) = √(5 - 2x). For square root functions, the expression under the radical must be non-negative (≥ 0) for real number outputs. We need 5 - 2x ≥ 0, which gives us 5 ≥ 2x, or x ≤ 5/2. The domain is all real numbers x such that x ≤ 5/2, written as (-∞, 5/2] or x ≤ 5/2. A common error is solving the inequality incorrectly or using the wrong inequality direction. When finding domains of radical functions, set up the inequality for the radicand ≥ 0 and solve carefully.

Question 13

A bacteria culture is modeled by the exponential function $P(t)=120\cdot(1.25)^t$ , where $t$ is time in hours. Which statement best describes the growth of the culture, and what is $P(4)$ ? Choose the option that correctly identifies the type of change and the value after 4 hours.

Multiply by $1.25$ each hour; $P(4)=292.97$ (correct answer)
Add $25$ each hour; $P(4)=220$
Add $30$ each hour; $P(4)=240$
Multiply by $0.75$ each hour; $P(4)=50.63$

Explanation: This question asks us to identify the growth pattern of an exponential function P(t) = 120·(1.25)^t and calculate P(4). This is an exponential function because the variable t appears in the exponent. In exponential growth, the quantity is multiplied by the base (1.25) each time period, not added to - so the culture multiplies by 1.25 each hour. To find P(4), we calculate: P(4) = 120·(1.25)^4 = 120·2.44140625 ≈ 292.97. The key error to avoid is treating exponential growth as linear (adding a constant each hour). When you see a function with the variable in the exponent, remember it represents multiplicative change.

Question 14

A bacteria culture is modeled by the function $P(t)=120\cdot(1.25)^t$ , where $t$ is the number of hours after the start. Which statement best describes the growth shown by this model, and what is the growth factor per hour?

Linear; adds $25$ each hour
Exponential; multiplies by $0.75$
Quadratic; adds $1.25t$
Exponential; multiplies by $1.25$ (correct answer)

Explanation: This question asks us to identify the function type and growth factor for P(t) = 120·(1.25)^t. The function has the form a·b^t where a = 120 and b = 1.25, which is the standard form of an exponential function. Since the base 1.25 > 1, this represents exponential growth where the population multiplies by 1.25 each hour. A common error is confusing the growth factor (1.25) with the growth rate (0.25 or 25%). When analyzing exponential models, look for the form a·b^t and identify b as the multiplicative factor per time unit.

Question 15

The curve shown is a rational function with a vertical asymptote at $x=2$ and a horizontal asymptote at $y=1$ . It passes through the point $(3,2)$ . Which equation matches these features?

$y=\dfrac{x-2}{x}$
$y=\dfrac{1}{x-2}$
$y=1+\dfrac{2}{x-2}$
$y=1+\dfrac{1}{x-2}$ (correct answer)

Explanation: The question asks which equation has a vertical asymptote at x = 2, a horizontal asymptote at y = 1, and passes through (3, 2). Writing a rational function as a constant plus a shifted reciprocal makes both features readable: y = 1 + 1/(x - 2) is undefined at x = 2, and as x grows the fraction approaches 0, leaving y approaching 1. Checking the point, 1 + 1/(3 - 2) = 1 + 1 = 2, so it fits. The version with 2/(x - 2) added to 1 has the right asymptotes but gives 1 + 2/1 = 3 at x = 3, so it misses the point. The plain 1/(x - 2) has its horizontal asymptote at y = 0, not 1. And (x - 2)/x puts the vertical asymptote at x = 0.

Question 16

A coordinate plane shows an exponential curve that passes through $(0,8)$ and decreases to the right, passing near $(1,4)$ and $(2,2)$ . Which equation best matches the graph?

$y=8(2)^x$
$y=8\left(\dfrac12\right)^x$ (correct answer)
$y=8-x$
$y=4\left(\dfrac12\right)^x$

Explanation: We need to identify an exponential curve passing through (0,8), (1,4), and (2,2). This is exponential decay since values are decreasing. From 8 to 4 is ×1/2, and from 4 to 2 is also ×1/2, confirming constant ratio. Testing y = 8(1/2)^x: at x = 0, y = 8(1/2)^0 = 8(1) = 8 ✓; at x = 1, y = 8(1/2)^1 = 8(1/2) = 4 ✓; at x = 2, y = 8(1/2)^2 = 8(1/4) = 2 ✓. The decay factor of 1/2 means the function value halves with each unit increase in x. Exponential decay has base between 0 and 1. When consecutive y-values show constant ratios less than 1, it indicates exponential decay.

Question 17

The function $p(x)=\dfrac{4}{x}+1$ is defined for real $x$ where possible. What is the range of $p(x)$ ?

$y\ne 0$
$y\ne 1$ (correct answer)
$y\ge 1$
$y\le 1$

Explanation: We need to find the range of p(x) = 4/x + 1, a rational function. The term 4/x can take any value except 0 (since x ≠ 0 for the function to be defined). As x approaches 0 from the positive side, 4/x → +∞; as x approaches 0 from the negative side, 4/x → -∞; as x → ±∞, 4/x → 0. Therefore, 4/x can be any real number except 0, and p(x) = 4/x + 1 can be any real number except 1. The range is y ≠ 1. The horizontal asymptote at y = 1 confirms this. For rational functions of the form a/x + k, the range excludes y = k.

Question 18

A coordinate plane shows the graph of $y=\sqrt{x-1}$ . What is the domain of this function as shown (real outputs only)?

$(-\infty,1]$
$[1,\infty)$ (correct answer)
$(1,\infty)$
$(-\infty,\infty)$

Explanation: This question asks for the domain of y = √(x-1) when considering only real outputs. This is a square root function, which requires the expression under the radical to be non-negative. Setting up the inequality: x - 1 ≥ 0, which gives x ≥ 1. The domain is all x-values greater than or equal to 1, written as [1,∞). The square bracket indicates that x = 1 is included since √(1-1) = √0 = 0 is defined. A common error is using parentheses instead of brackets, which would exclude the endpoint. For radical functions with even roots, always use ≥ (not >) for the domain restriction.

Question 19

A coordinate plane shows a curve starting at $(4,0)$ and increasing to the right, passing through $(5,1)$ and $(8,2)$ . Which equation best matches the graph?

$y=\sqrt{x-4}$ (correct answer)
$y=|x-4|$
$y=(x-4)^2$
$y=\dfrac{1}{x-4}$

Explanation: We need to match an equation to a curve starting at (4,0) and passing through (5,1) and (8,2). This is a square root function since it starts at a point and increases at a decreasing rate. Testing y = √(x-4): at x = 4, y = √(4-4) = √0 = 0 ✓; at x = 5, y = √(5-4) = √1 = 1 ✓; at x = 8, y = √(8-4) = √4 = 2 ✓. The function starts at x = 4 because that's where the expression under the radical becomes zero. Square root functions have a starting point and increase gradually. The characteristic shape and the fact it starts at (4,0) rather than continuing left confirms this is a radical function.

Question 20

For the function $f(x)=|2x-7|$ , what is the value of $f(1)$ ?

$-5$
$5$ (correct answer)
$-9$
$9$

Explanation: We need to find f(1) for the absolute value function f(x) = |2x - 7|. This is an absolute value function, which creates a V-shaped graph. To find f(1), substitute x = 1: f(1) = |2(1) - 7| = |2 - 7| = |-5| = 5. The absolute value operation takes the distance from zero, so |-5| = 5. A common error is forgetting to apply the absolute value and leaving the answer as -5. When working with absolute value functions, always remember the final step of taking the positive value.