PSAT Math Quiz: Solving Nonlinear Functions

20 questions · exam conditions

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Solving Nonlinear FunctionsQuestion 1 of 20

If $f(x)=2^{x+1}$ , what is $f(3)$ ? Be careful: the exponent is $x+1$ , not just $x$ .

16

8

32

12

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PSAT Math Quiz

PSAT Math Quiz: Solving Nonlinear Functions

Practice Solving Nonlinear Functions in PSAT 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 PSAT 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

If $f(x)=2^{x+1}$ , what is $f(3)$ ? Be careful: the exponent is $x+1$ , not just $x$ .

$16$ (correct answer)
$8$
$32$
$12$

Explanation: The question asks for

f(3)

where

f(x)=2^{x+1}

. This is exponential, nonlinear with base 2. Substitute:

2^{3+1}=2^4=16

. The exponent is

x+1

, not

x

. A common error is forgetting +1, getting 8 like B. Another is miscalculating

2^3=8

. Apply the exponent correctly to avoid confusing with base changes.

Question 2

A store's online sales follow $S(t)=150\cdot 1.2^t$ , where $t$ is weeks. Which best describes the growth rate? Interpret $1.2$ as a weekly multiplier.

20% increase/week (correct answer)
120% increase/week
20% decrease/week
1.2% increase/week

Explanation: This question asks for the best description of the growth rate in the given exponential sales model. The function is an exponential growth function since the base 1.2 is greater than 1, distinguishing it from decay functions where the base is between 0 and 1. To find the growth rate, interpret the base 1.2 as a multiplier, meaning sales increase by a factor of 1.2 each week, which corresponds to a 20% increase because 1.2 = 1 + 0.2. Subtracting 1 from the base and converting to a percentage gives the rate directly. A common error is confusing the multiplier with the percentage increase, such as thinking 1.2 means 120% increase, which would actually be a multiplier of 2.2. When dealing with exponential models, always express the growth rate as (base - 1) × 100% for percentage interpretation.

Question 3

For the rational function $r(x)=\dfrac{2x-1}{x-4}$ , what is $r(6)$ ? Substitute carefully and simplify the fraction.

$\tfrac{11}{2}$ (correct answer)
$\tfrac{1}{11}$
$\tfrac{11}{10}$
$\tfrac{10}{11}$

Explanation: The question asks for r(6) where

r(x) = \frac{2x - 1}{x - 4}

. This is a rational function, nonlinear and evaluated by substitution. Plug in x=6:

\frac{12 - 1}{6 - 4} = \frac{11}{2}

. Simplify the fraction after substituting. A common error is mis-substituting like getting

\frac{1}{11}

. Another is confusing numerator and denominator. For rationals, substitute carefully and simplify to avoid arithmetic errors.

Question 4

The function $g(x)=\dfrac{x}{x+5}$ is nonlinear. What is $g(-5)$ ? Identify what happens when the denominator becomes 0.

$-1$
$0$
Undefined (correct answer)
$1$

Explanation: The question requires finding

g(-5)

for

g(x) = \dfrac{x}{x+5}

, noting it's nonlinear. This is a rational function, nonlinear with a possible asymptote. At

x=-5

, denominator

-5 + 5 = 0

, so undefined. Division by zero makes it undefined, not a real number. Matches choice C. A common error is attempting to compute it as

-5/0

= some value. Strategy: Check denominator for zero before evaluating rational functions to identify undefined points.

Question 5

A coordinate plane shows a rational function with vertical asymptote at $x=-2$ and horizontal asymptote at $y=1$ . The graph passes through $(-1,3)$ . Which equation matches these features?

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

Explanation: The question asks for the equation matching a rational function with vertical asymptote x=-2, horizontal y=1, through (-1,3). This is rational, nonlinear with asymptotes. Form y = a/(x+2) + 1; at x=-1, a/1 +1=3 so a=2, giving y=2/(x+2)+1. Verify asymptotes and point. A key error is shifting wrong like B or numerator like C. Linear D lacks asymptotes. Use point to solve for coefficient, distinguishing from pure inverses.

Question 6

Solve for all real values of $x$ that satisfy $\sqrt{2x+3}=x-1$ .

$x=2\pm\sqrt6$ (both values)
$x=2-\sqrt6$ only
$x=2+\sqrt6$ only (correct answer)
No real solutions

Explanation: When solving equations involving square roots, you must be careful about extraneous solutions that can emerge during the solving process. To solve

\sqrt{2x+3}=x-1

, start by noting that the left side requires

2x+3 \geq 0

, so

x \geq -\frac{3}{2}

. Additionally, since

\sqrt{2x+3} \geq 0

, we need

x-1 \geq 0

, meaning

x \geq 1

. Square both sides to eliminate the square root:

2x+3 = (x-1)^2

. Expanding the right side gives

2x+3 = x^2-2x+1

. Rearranging:

0 = x^2-4x-2

. Using the quadratic formula:

x = \frac{4 \pm \sqrt{16+8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}

. Now you must check both solutions in the original equation. For

x = 2-\sqrt{6}

: Since

\sqrt{6} \approx 2.45

, this gives

x \approx -0.45

, which violates our requirement that

x \geq 1

. For

x = 2+\sqrt{6} \approx 4.45

: This satisfies

x \geq 1

and checking shows it works in the original equation. Choice A incorrectly includes both values without checking for extraneous solutions. Choice B selects only the extraneous solution

x = 2-\sqrt{6}

. Choice D incorrectly concludes there are no solutions when there is actually one valid solution. Always check your solutions when solving radical equations—squaring both sides can introduce extraneous solutions that don't satisfy the original equation's domain restrictions.

Question 7

Which of the following gives all real solutions to $|2x-7|=x+1$ ?

$x=8 \text{ only}$
$x=-4 \text{ or } x=8$
$x=2 \text{ only}$
$x=2 \text{ or } x=8$ (correct answer)

Explanation: When you encounter an absolute value equation, remember that the expression inside the absolute value bars can be either positive or negative, leading to two separate cases to solve. For

|2x-7|=x+1

, you need to consider both possibilities: Case 1:

2x-7=x+1

(when the expression inside is positive) Solving:

2x-x=1+7

, so

x=8

Case 2:

2x-7=-(x+1)

(when the expression inside is negative)
Solving:

2x-7=-x-1

, which gives

3x=6

, so

x=2

Always check your solutions by substituting back into the original equation:

For $x=8$ : $|2(8)-7|=|9|=9$ and $8+1=9$ ✓
For $x=2$ : $|2(2)-7|=|-3|=3$ and $2+1=3$ ✓

Both solutions work, so the answer is D. Now for the wrong choices: A only gives

x=8

, missing the second valid solution. B incorrectly includes

x=-4

—if you substitute this back, you get

|2(-4)-7|=15

but

-4+1=-3

, which don't equal each other. C only gives

x=2

, missing the first valid solution. Strategy tip: For absolute value equations, always solve both cases systematically, then verify each solution in the original equation. Students often forget to check their work or miss one of the two cases, leading to incomplete answers.

Question 8

A bacteria culture starts with 200 cells and doubles every 3 hours. The number of cells after $t$ hours is modeled by a nonlinear function. Which equation represents the number of cells after $t$ hours? Be careful to use multiplicative growth rather than adding a constant amount each hour.

$N(t)=200+2t$
$N(t)=200\cdot 2^{t/3}$ (correct answer)
$N(t)=2\cdot 200^{t/3}$
$N(t)=200\cdot (t/3)^2$

Explanation: The question asks for the equation modeling the number of bacteria cells after t hours, starting with 200 cells and doubling every 3 hours. This is an exponential growth function, a type of nonlinear function where the quantity multiplies by a constant factor over time. To find the model, recognize that doubling every 3 hours means the growth factor is 2 raised to the power of t/3, so N(t) = 200 * 2^{t/3}. Verify by checking at t=3: N(3) = 200 * 2^1 = 400, which is double the initial amount. A key error is choosing linear growth like choice A, which adds a constant instead of multiplying, failing to capture exponential doubling. Another mistake is swapping the base and initial value as in choice C, which grows too rapidly. For growth problems, distinguish exponential from linear by checking if changes are multiplicative rather than additive.

Question 9

A coordinate plane shows the graph of $y=|x-1|-2$ . Which statement correctly identifies a key feature of this function?

Vertex at $(1,-2)$ (correct answer)
Vertex at $(-1,2)$
Vertex at $(1,2)$
Vertex at $(-2,1)$

Explanation: This question asks about the vertex of an absolute value function. The function y = |x-1|-2 is in the form y = |x-h|+k, where (h,k) represents the vertex. From the given equation, we can identify h = 1 (note the sign: x-1 means we shift right by 1) and k = -2. Therefore, the vertex is at (1,-2), which is the point where the V-shaped graph changes direction. A common error is to misinterpret the signs, thinking x-1 means the vertex is at x = -1. Remember that |x-h| shifts the graph h units to the right when h is positive.

Question 10

A car's value is $\$ 18{,}000$ when purchased and then depreciates by 15% each year. What is the value after 2 years, according to an exponential model? Round to the nearest dollar.

$\$ 15{,}300$
$\$ 13{,}005$ (correct answer)
$\$ 12{,}600$
$\$ 20{,}700$

Explanation: The question asks for the car's value after 2 years, starting at $18,000 depreciating 15% yearly. This is exponential decay, nonlinear with factor 0.85. Calculate 18000 * 0.85^2 = 18000 * 0.7225 = 13005. Round to nearest dollar. A common error is subtracting like getting 12600 in C. Another is wrong calculation like A. Use decay formula and verify multiplication to avoid linear mistakes.

Question 11

A bacteria culture starts with 500 cells and doubles every 3 hours. The number of cells after $t$ hours is modeled by $N(t)=500\cdot 2^{t/3}$ . Which expression gives the number of cells after 9 hours, and what is its value?

$500\cdot 2^{3}=4000$
$500\cdot 2^{9}=256000$
$500\cdot 9^{2}=40500$
$500\cdot 2^{9/3}=4000$ (correct answer)

Explanation: This question asks us to find the number of bacteria cells after 9 hours using the given exponential growth model. The function N(t) = 500·2^(t/3) is an exponential function where the bacteria population doubles every 3 hours. To find the number of cells after 9 hours, we substitute t = 9 into the formula: N(9) = 500·2^(9/3) = 500·2^3 = 500·8 = 4000. A common error is to use 2^9 instead of 2^(9/3), which would give an incorrect answer of 256,000 cells. When dealing with exponential growth problems, always check that the exponent correctly reflects the given time period and growth rate.

Question 12

What is the solution to the rational equation $\dfrac{x+2}{x-1}=3$ ?

$x=3$
$x=\dfrac52$ (correct answer)
$x=\dfrac32$
$x=1$ (not permitted)

Explanation: When you encounter a rational equation like this one, your goal is to isolate the variable by eliminating the fraction. The most efficient approach is to multiply both sides by the denominator. Starting with

\frac{x+2}{x-1}=3

, multiply both sides by

(x-1)

(x-1) \cdot \frac{x+2}{x-1} = 3 \cdot (x-1)

The left side simplifies to just

(x+2)

, giving you:

x+2 = 3(x-1)

Distribute the 3 on the right side:

x+2 = 3x-3

Solve for

x

by collecting like terms:

x+2 = 3x-3

2+3 = 3x-x

5 = 2x

x = \frac{5}{2}

This confirms answer choice B is correct. Let's check why the other options are wrong. Choice A (

x=3

) would give you

\frac{3+2}{3-1} = \frac{5}{2} = 2.5

, not 3. Choice C (

x=\frac{3}{2}

) would yield

\frac{\frac{3}{2}+2}{\frac{3}{2}-1} = \frac{\frac{7}{2}}{\frac{1}{2}} = 7

, not 3. Choice D (

x=1

) is explicitly marked as "not permitted" because it makes the denominator zero, creating an undefined expression. Strategy tip: Always check that your solution doesn't make any denominator equal to zero. For rational equations, certain values are automatically excluded from the domain, and recognizing these restrictions will help you avoid trap answers on the PSAT.

Question 13

A coordinate plane shows a hyperbola-like graph with vertical asymptote at $x=1$ and horizontal asymptote at $y=0$ . The graph passes through $(2,3)$ . Which equation matches the graph?

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

Explanation: The question asks for the equation matching a hyperbola-like graph with vertical asymptote at x=1, horizontal at y=0, passing through (2,3). This is a rational function, nonlinear with asymptotes indicating poles and degree differences. The form y = a/(x - 1) fits: at x=2, a/1 = 3 so a=3, giving y=3/(x-1). Verify asymptotes: vertical at x=1, horizontal y=0 since degrees match with constant numerator. A key error is shifting incorrectly like B or using linear forms like C. Inverse variation might confuse, but check points. For asymptotes, identify shifts to differentiate rationals from exponentials.

Question 14

If $f(x)=\dfrac{3x}{x-1}$ , which statement describes the domain? Choose the correct restriction on $x$ so the function is defined.

All real $x$
All real $x\ne 0$
All real $x\ne 1$ (correct answer)
All real $x\ge 1$

Explanation: The question asks for the domain description of f(x) = 3x/(x - 1). This is a rational function, nonlinear undefined where denominator is zero. Set x - 1 ≠ 0, so x ≠ 1, domain all reals except 1. Matches choice C. Other restrictions like x ≥ 1 are for square roots. A key error is assuming no restrictions like polynomials. Strategy: Identify denominator zeros to define domains for rational functions, excluding those points.

Question 15

A car worth $\$ 18{,}000 $depreciates by 12% each year. If$ V(t) $is the value after$ t$ years, which equation represents this situation?

$V(t)=18000(1.12)^t$
$V(t)=18000(0.88)^t$ (correct answer)
$V(t)=18000-0.12t$
$V(t)=18000-0.12(18000)^t$

Explanation: This question asks for an exponential decay model for a depreciating car. When something depreciates by 12% each year, it retains 88% (or 0.88) of its value. The exponential decay formula is V(t) = V₀(1-r)^t, where V₀ is the initial value and r is the decay rate. With V₀ = 18000 and r = 0.12, we get V(t) = 18000(1-0.12)^t = 18000(0.88)^t. A common mistake is to use 1.12 instead of 0.88, which would represent growth rather than decay. Remember that depreciation means the value decreases, so the base of the exponential must be less than 1.

Question 16

Evaluate the nonlinear function $f(x)=|3x-7|$ at $x=1$ . Pay attention to the absolute value bars when the expression inside is negative.

$-4$
$4$ (correct answer)
$10$
$-10$

Explanation: The question asks to evaluate

f(x) = |3x - 7|

x=1

. This is an absolute value function, nonlinear and piecewise depending on the sign inside. At

x=1

3*1 - 7 = -4

, so

| -4 | = 4

. The expression inside is negative, so it flips to positive. A key error is ignoring the absolute value and getting

-4

like A. Another mistake is miscalculating the inside. When evaluating absolute values, check the sign inside to distinguish from linear evaluations.

Question 17

A coordinate plane shows a curve with vertex at $(1,0)$ opening upward and passing through $(0,1)$ and $(2,1)$ . Which equation matches the graph?

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

Explanation: The question seeks the equation for a graph with vertex at (1,0), opening upward through (0,1) and (2,1). This is a quadratic function, nonlinear and parabolic. Vertex form suggests y = (x - 1)²: at x=0, 1; x=2, 1; x=1, 0, matches. Absolute value would be V-shaped, not parabolic. Choice B fits. Common error: Choosing shifted forms without testing points. Strategy: Use vertex and symmetry points to fit quadratics, distinguishing from absolute value or square root.

Question 18

A radioactive sample has mass $M(t)=50(0.8)^t$ grams after $t$ days. What is $M(3)$ ? Compute using the exponential decay factor.

$25.6$ (correct answer)
$32$
$40$
$64$

Explanation: The question asks for M(3) in the decay model M(t) = 50 (0.8)^t. This is an exponential function, nonlinear modeling radioactive decay. Compute (0.8)^3 = 0.512, then 50 * 0.512 = 25.6. This matches choice A. Other values like 32 might ignore the exponent. A key error is multiplying instead of exponentiating the base. Strategy: Use order of operations for exponents in decay models to avoid miscalculation.

Question 19

A coordinate plane shows a curve with a vertical asymptote at $x=0$ and it passes through $(1,2)$ and $(2,1)$ . The graph is in quadrants I and III and decreases as $x$ increases in quadrant I. Which function best matches?

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

Explanation: The question asks for the function matching a curve with vertical asymptote x=0, through (1,2) and (2,1), in quadrants I and III, decreasing in QI. This is rational like inverse, nonlinear with y=2/x fitting points and asymptotes. Verify: at x=1,2; x=2,1; negative for x<0. A key error is linear like B or exponential like C. Square root D is only positive. Check quadrants and points to identify rational over exponential.

Question 20

A coordinate plane shows a curve that is symmetric about the y-axis, passes through $(0,0)$ , and has points $(2,4)$ and $(-2,4)$ . Which equation matches the graph?

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

Explanation: The question requires the equation for a graph symmetric about the y-axis, passing through (0,0), (2,4), and (-2,4). This is a quadratic function, nonlinear and even, like y = x². Symmetry about y-axis means f(-x) = f(x); testing y = x²: at x=0, 0; x=2, 4; x=-2, 4, matches. Absolute value y=|x| passes (0,0) but (2,2) not 4. This is choice A. Common error: Selecting absolute value without verifying points. Strategy: Check symmetry and plug in points to distinguish even quadratics from other symmetric nonlinear functions like absolute value.