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College Algebra Quiz

College Algebra Quiz: Function Notation Evaluate And Interpret

Practice Function Notation Evaluate And Interpret in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 1

0 of 1 answered

Consider the function f(x)=∣x−3∣+2f(x) = |x - 3| + 2f(x)=∣x−3∣+2. If f(a)=f(b)f(a) = f(b)f(a)=f(b) where a≠ba \neq ba=b, and both aaa and bbb are in the domain [−1,7][-1, 7][−1,7], which of the following must be true?

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What this quiz covers

This quiz focuses on Function Notation Evaluate And Interpret, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

Consider the function f(x)=∣x−3∣+2f(x) = |x - 3| + 2f(x)=∣x−3∣+2. If f(a)=f(b)f(a) = f(b)f(a)=f(b) where a≠ba \neq ba=b, and both aaa and bbb are in the domain [−1,7][-1, 7][−1,7], which of the following must be true?

  1. a+b=6a + b = 6a+b=6 and one of a,ba, ba,b is less than 3 while the other is greater than 3 (correct answer)
  2. a+b=6a + b = 6a+b=6 and both aaa and bbb are greater than or equal to 3
  3. a+b=3a + b = 3a+b=3 and one of a,ba, ba,b is less than 3 while the other is greater than 3
  4. ∣a−3∣=∣b−3∣|a - 3| = |b - 3|∣a−3∣=∣b−3∣ but a+ba + ba+b could be any value in the given domain

Explanation: The function f(x)=∣x−3∣+2f(x) = |x - 3| + 2f(x)=∣x−3∣+2 has its vertex at x=3x = 3x=3, where it achieves its minimum value of f(3)=2f(3) = 2f(3)=2. For x<3x < 3x<3, we have f(x)=−(x−3)+2=−x+5f(x) = -(x - 3) + 2 = -x + 5f(x)=−(x−3)+2=−x+5. For x≥3x \geq 3x≥3, we have f(x)=(x−3)+2=x−1f(x) = (x - 3) + 2 = x - 1f(x)=(x−3)+2=x−1. If f(a)=f(b)f(a) = f(b)f(a)=f(b) with a≠ba \neq ba=b, then we need ∣a−3∣+2=∣b−3∣+2|a - 3| + 2 = |b - 3| + 2∣a−3∣+2=∣b−3∣+2, which simplifies to ∣a−3∣=∣b−3∣|a - 3| = |b - 3|∣a−3∣=∣b−3∣. This occurs when either a−3=b−3a - 3 = b - 3a−3=b−3 (giving a=ba = ba=b, which contradicts a≠ba \neq ba=b) or a−3=−(b−3)a - 3 = -(b - 3)a−3=−(b−3), which gives a−3=−b+3a - 3 = -b + 3a−3=−b+3, so a+b=6a + b = 6a+b=6. For this to happen with a≠ba \neq ba=b, one value must be on the left side of the vertex (<3< 3<3) and the other on the right side (>3> 3>3). For example, if a=1a = 1a=1, then f(1)=2+2=4f(1) = 2 + 2 = 4f(1)=2+2=4, and we need f(b)=4f(b) = 4f(b)=4. From b−1=4b - 1 = 4b−1=4, we get b=5b = 5b=5. Indeed, a+b=1+5=6a + b = 1 + 5 = 6a+b=1+5=6. Choice D is incorrect because while ∣a−3∣=∣b−3∣|a - 3| = |b - 3|∣a−3∣=∣b−3∣ is necessary, it doesn't capture the full constraint that a+b=6a + b = 6a+b=6.