PSAT Math Quiz: Solving Nonlinear Functions

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.

Question 1

1. $16$ (correct answer)
2. $8$
3. $32$
4. $12$

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

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

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 3

1. $-1$
2. $0$
3. Undefined (correct answer)
4. $1$

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 4

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

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 5

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

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 6

1. Vertex at $(1,-2)$ (correct answer)
2. Vertex at $(-1,2)$
3. Vertex at $(1,2)$
4. 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 7

1. \15{,}300$
2. \13{,}005$ (correct answer)
3. \12{,}600$
4. \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 8

1. $500\cdot 2^{3}=4000$
2. $500\cdot 2^{9}=256000$
3. $500\cdot 9^{2}=40500$
4. $500\cdot 2^{9/3}=4000$ (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 9

1. $x=3$
2. $x=\dfrac52$ (correct answer)
3. $x=\dfrac32$
4. $x=1$ (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 10

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

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 11

1. All real $x$
2. All real $x\ne 0$
3. All real $x\ne 1$ (correct answer)
4. 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 12

1. $V(t)=18000(1.12)^t$
2. $V(t)=18000(0.88)^t$ (correct answer)
3. $V(t)=18000-0.12t$
4. $V(t)=18000-0.12(18000)^t$

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 13

1. $-4$
2. $4$ (correct answer)
3. $10$
4. $-10$

Explanation: The question asks to evaluate $f(x) = |3x - 7|$ at $x=1$. 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 14

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

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 15

1. $25.6$ (correct answer)
2. $32$
3. $40$
4. $64$

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 16

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

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 17

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

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.

Question 18

1. $C(t)=80(1.25)^t$
2. $C(t)=80(0.75)^t$ (correct answer)
3. $C(t)=80-25t$
4. $C(t)=80(0.25)^t$

Explanation: The question asks for the equation modeling medicine concentration starting at 80 mg/L, decreasing by 25% hourly. This is exponential decay, nonlinear with factor (1 - 0.25)=0.75. Thus, C(t)=80 * (0.75)^t. Verify: after 1 hour, 80*0.75=60. A key error is using growth like A or linear subtraction like C. Choice D has wrong factor. For decay, calculate the retention factor to avoid confusing with growth models.

Question 19

1. $y=10\cdot 2^x$
2. $y=10\left(\tfrac12\right)^x$ (correct answer)
3. $y=5\left(\tfrac12\right)^x$
4. $y=10-5x$

Explanation: This question asks for the equation that matches the described exponential decay curve on the coordinate plane. The function is exponential decay because it decreases over time, distinguished from growth where the base exceeds 1 or linear functions. The y-intercept is 10, so when x=0, y=10, and it passes through (1,5), so substitute x=1, y=5 into y=10*(1/2)^x to verify: 10*(0.5)=5, which fits. For other choices, y=102^x gives 20 at x=1, and the linear option y=10-5x passes through (1,5) but is not exponential. A common error is selecting a decay function with the wrong initial value or base, like y=5(1/2)^x, which starts at 5. When graphs are described, plug in given points to test equations quickly.

Question 20

1. $A(t)=500(1.06)^t$ (correct answer)
2. $A(t)=500+0.06t$
3. $A(t)=500(0.06)^t$
4. $A(t)=500(1.6)^t$

Explanation: The question asks for the equation modeling a savings account starting at $500 and growing by 6% yearly. This is exponential growth, a nonlinear function with multiplication by (1 + rate). The model is A(t) = 500 * (1.06)^t, where 1.06 is the growth factor. Verify: after 1 year, 500*1.06=530. A common error is using addition like B or decay like C. Choice D has incorrect base. For percent growth, use exponential form and check the base to avoid confusing with linear models.