Middle School Math Quiz: Evaluating Functions
5 questions · exam conditions
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Evaluating FunctionsQuestion 1 of 5

Maria's monthly phone bill is calculated using the function B(m)=45+0.15mB(m) = 45 + 0.15m, where mm represents the number of minutes used beyond her basic plan.

If Maria's bill last month was $63, how much would her bill be this month if she uses 40 fewer minutes than last month?

$57.00
$59.25
$61.50
$51.00
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Middle School Math Quiz

Middle School Math Quiz: Evaluating Functions

Practice Evaluating Functions in Middle School 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 Evaluating Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School 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

Maria's monthly phone bill is calculated using the function B(m)=45+0.15mB(m) = 45 + 0.15m, where mm represents the number of minutes used beyond her basic plan.

If Maria's bill last month was $63, how much would her bill be this month if she uses 40 fewer minutes than last month?

  1. $57.00 (correct answer)
  2. $59.25
  3. $61.50
  4. $51.00
Explanation: First, find last month's minutes: 45+0.15m=6345 + 0.15m = 63, so 0.15m=180.15m = 18 and m=120m = 120. This month she uses 120−40=80120 - 40 = 80 minutes beyond basic. So B(80)=45+0.15(80)=45+12=57B(80) = 45 + 0.15(80) = 45 + 12 = 57. Choice B uses 95 minutes (adding instead of subtracting 40). Choice C uses 110 minutes (subtracting only 10). Choice D subtracts $6 directly from the original bill.

Question 2

The function g(x)=x2−4x+1g(x) = x^2 - 4x + 1 models the height of a ball in feet after xx seconds. At what time(s) other than t=1t = 1 second is the ball at the same height as it was at t=1t = 1 second?

  1. t=2t = 2 seconds only
  2. t=3t = 3 seconds only (correct answer)
  3. t=0t = 0 and t=4t = 4 seconds
  4. t=1t = 1 and t=3t = 3 seconds
Explanation: First find g(1)=1−4+1=−2g(1) = 1 - 4 + 1 = -2. Then solve x2−4x+1=−2x^2 - 4x + 1 = -2, which gives x2−4x+3=0x^2 - 4x + 3 = 0. Factoring: (x−1)(x−3)=0(x-1)(x-3) = 0, so x=1x = 1 or x=3x = 3. Since we want times other than t=1t = 1, the answer is t=3t = 3. Choice A uses the axis of symmetry x=2x = 2. Choice C solves x2−4x+1=1x^2 - 4x + 1 = 1. Choice D includes the original time t=1t = 1.

Question 3

A piecewise function is defined as f(x)={x+5if x≤23x−1if x>2f(x) = \begin{cases} x + 5 & \text{if } x \leq 2 \\ 3x - 1 & \text{if } x > 2 \end{cases} . What is the value of f(f(1))f(f(1))?

  1. 5
  2. 8
  3. 17 (correct answer)
  4. 11
Explanation: First find f(1)f(1). Since 1≤21 \leq 2, use the first piece: f(1)=1+5=6f(1) = 1 + 5 = 6. Then find f(6)f(6). Since 6>26 > 2, use the second piece: f(6)=3(6)−1=17f(6) = 3(6) - 1 = 17. Choice A is just f(1)−1f(1) - 1. Choice B uses the wrong piece for f(6)f(6): 6+5−3=86 + 5 - 3 = 8. Choice D results from using 3(1)−1=23(1) - 1 = 2 then 2+5+4=112 + 5 + 4 = 11.

Question 4

A function ff is defined by f(x)=3x−7f(x) = 3x - 7. If f(a)=8f(a) = 8, what is the value of f(2a)f(2a)?

  1. 16
  2. 23 (correct answer)
  3. 9
  4. 30
Explanation: First, find aa by solving f(a)=8f(a) = 8: 3a−7=83a - 7 = 8, so 3a=153a = 15 and a=5a = 5. Then f(2a)=f(10)=3(10)−7=23f(2a) = f(10) = 3(10) - 7 = 23. Choice A results from calculating 2⋅8=162 \cdot 8 = 16. Choice C comes from finding aa correctly but calculating f(a+a)f(a+a) incorrectly. Choice D results from f(2)⋅f(a)=−1⋅8f(2) \cdot f(a) = -1 \cdot 8.

Question 5

Given f(x)=2x2−x+3f(x) = 2x^2 - x + 3, which expression represents f(x+1)−f(x)f(x+1) - f(x)?

  1. 4x2+3x+24x^2 + 3x + 2
  2. 4x+34x + 3
  3. 2x+12x + 1
  4. 4x+14x + 1 (correct answer)
Explanation: When you encounter function composition problems like this, you're working with substitution and algebraic simplification. The key is to carefully substitute the new input into the original function, then perform the required operations. To find f(x+1)−f(x)f(x+1) - f(x), you first need to calculate f(x+1)f(x+1) by substituting (x+1)(x+1) everywhere you see xx in the original function. Starting with f(x)=2x2−x+3f(x) = 2x^2 - x + 3: f(x+1)=2(x+1)2−(x+1)+3f(x+1) = 2(x+1)^2 - (x+1) + 3 Expanding (x+1)2=x2+2x+1(x+1)^2 = x^2 + 2x + 1: f(x+1)=2(x2+2x+1)−(x+1)+3f(x+1) = 2(x^2 + 2x + 1) - (x+1) + 3 f(x+1)=2x2+4x+2−x−1+3f(x+1) = 2x^2 + 4x + 2 - x - 1 + 3 f(x+1)=2x2+3x+4f(x+1) = 2x^2 + 3x + 4 Now subtract f(x)f(x): f(x+1)−f(x)=(2x2+3x+4)−(2x2−x+3)f(x+1) - f(x) = (2x^2 + 3x + 4) - (2x^2 - x + 3) =2x2+3x+4−2x2+x−3= 2x^2 + 3x + 4 - 2x^2 + x - 3 =4x+1= 4x + 1 Choice A (4x2+3x+24x^2 + 3x + 2) likely comes from forgetting to subtract f(x)f(x) completely. Choice B (4x+34x + 3) results from arithmetic errors in the constant terms. Choice C (2x+12x + 1) suggests mistakes when combining like terms, possibly missing some xx terms during subtraction. Remember that difference problems like f(x+1)−f(x)f(x+1) - f(x) often simplify dramatically because many terms cancel out. Always distribute carefully when expanding, and double-check your arithmetic when combining like terms—the quadratic terms should disappear entirely in this type of problem.