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College Algebra

College Algebra Quiz: Graphing From Parent Functions

Practice Graphing From Parent Functions in College Algebra 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 Graphing From Parent Functions, 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.

Question 1

If f(x)=x3f(x) = x^3f(x)=x3 and g(x)=12(x+1)3−4g(x) = \frac{1}{2}(x + 1)^3 - 4g(x)=21​(x+1)3−4, then g(x)g(x)g(x) can be described as f(x)f(x)f(x) after which transformations?

  1. Horizontal compression by factor 12\frac{1}{2}21​, shift left 1 unit, shift down 4 units
  2. Vertical compression by factor 12\frac{1}{2}21​, shift left 1 unit, shift down 4 units
  3. Vertical compression by factor 12\frac{1}{2}21​, shift right 1 unit, shift down 4 units
  4. Horizontal compression by factor 12\frac{1}{2}21​, shift right 1 unit, shift up 4 units
Explanation: Starting with f(x)=x3f(x) = x^3f(x)=x3: The term (x+1)3(x + 1)^3(x+1)3 represents a horizontal shift left 1 unit. The coefficient 12\frac{1}{2}21​ represents a vertical compression by factor 12\frac{1}{2}21​. The −4-4−4 represents a vertical shift down 4 units. So g(x)=12(x+1)3−4g(x) = \frac{1}{2}(x + 1)^3 - 4g(x)=21​(x+1)3−4. Choice A confuses vertical and horizontal compression. Choice C shifts right instead of left. Choice D has multiple errors: wrong compression type, wrong horizontal direction, and wrong vertical direction.

Question 2

Consider the function h(x)=4−2x+3h(x) = \sqrt{4 - 2x} + 3h(x)=4−2x​+3. This function can be obtained by transforming the parent function f(x)=xf(x) = \sqrt{x}f(x)=x​. What is the domain of h(x)h(x)h(x)?

  1. (−∞,2](-\infty, 2](−∞,2]
  2. [2,∞)[2, \infty)[2,∞)
  3. [−2,∞)[-2, \infty)[−2,∞)
  4. (−∞,−2](-\infty, -2](−∞,−2]
Explanation: For h(x)=4−2x+3h(x) = \sqrt{4 - 2x} + 3h(x)=4−2x​+3 to be defined, we need 4−2x≥04 - 2x \geq 04−2x≥0. Solving: 4−2x≥0⇒4≥2x⇒2≥x4 - 2x \geq 0 \Rightarrow 4 \geq 2x \Rightarrow 2 \geq x4−2x≥0⇒4≥2x⇒2≥x, so x≤2x \leq 2x≤2. The domain is (−∞,2](-\infty, 2](−∞,2]. Choice B reverses the inequality. Choice C uses x≥−2x \geq -2x≥−2, which comes from incorrectly solving 4−2x≥04 - 2x \geq 04−2x≥0 as −2x≥−4-2x \geq -4−2x≥−4 without flipping the inequality sign. Choice D combines both errors.

Question 3

The parent function f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ is transformed to create h(x)=−2x+3+1h(x) = \frac{-2}{x+3} + 1h(x)=x+3−2​+1. What are the equations of the vertical and horizontal asymptotes of h(x)h(x)h(x)?

  1. Vertical: x=3x = 3x=3, Horizontal: y=1y = 1y=1
  2. Vertical: x=−3x = -3x=−3, Horizontal: y=−1y = -1y=−1
  3. Vertical: x=3x = 3x=3, Horizontal: y=−1y = -1y=−1
  4. Vertical: x=−3x = -3x=−3, Horizontal: y=1y = 1y=1
Explanation: When working with rational function transformations, you need to identify how the parent function f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ has been shifted to find the new asymptotes. The transformed function h(x)=−2x+3+1h(x) = \frac{-2}{x+3} + 1h(x)=x+3−2​+1 shows two key transformations from the parent function. The vertical asymptote occurs where the denominator equals zero, so you set x+3=0x + 3 = 0x+3=0, which gives x=−3x = -3x=−3. The horizontal asymptote is determined by the constant term added outside the fraction, which is +1+1+1, so the horizontal asymptote is y=1y = 1y=1. Looking at the incorrect choices: Choice A places the vertical asymptote at x=3x = 3x=3, which incorrectly assumes the shift is in the positive direction rather than recognizing that x+3=0x + 3 = 0x+3=0 means x=−3x = -3x=−3. Choice B correctly identifies the vertical asymptote as x=−3x = -3x=−3 but incorrectly places the horizontal asymptote at y=−1y = -1y=−1, confusing the sign of the vertical shift. Choice C combines both errors—wrong vertical asymptote and wrong horizontal asymptote. The correct answer is D: the vertical asymptote is x=−3x = -3x=−3 and the horizontal asymptote is y=1y = 1y=1. Study tip: For rational function transformations in the form ax−h+k\frac{a}{x-h} + kx−ha​+k, remember that the vertical asymptote is x=hx = hx=h (set the denominator to zero) and the horizontal asymptote is y=ky = ky=k (the vertical shift). Watch the signs carefully—x+3x + 3x+3 means h=−3h = -3h=−3, not +3+3+3.

Question 4

Which of the following functions represents a horizontal stretch of f(x)=(x−1)2+2f(x) = (x-1)^2 + 2f(x)=(x−1)2+2 by a factor of 3?

  1. g(x)=3(x−1)2+2g(x) = 3(x-1)^2 + 2g(x)=3(x−1)2+2
  2. g(x)=(3x−1)2+2g(x) = (3x-1)^2 + 2g(x)=(3x−1)2+2
  3. g(x)=(x3−1)2+2g(x) = \left(\frac{x}{3}-1\right)^2 + 2g(x)=(3x​−1)2+2
  4. g(x)=(x−13)2+2g(x) = \left(\frac{x-1}{3}\right)^2 + 2g(x)=(3x−1​)2+2
Explanation: A horizontal stretch by factor 3 means replacing xxx with x3\frac{x}{3}3x​ in the function. Starting with f(x)=(x−1)2+2f(x) = (x-1)^2 + 2f(x)=(x−1)2+2, we get g(x)=(x3−1)2+2g(x) = \left(\frac{x}{3}-1\right)^2 + 2g(x)=(3x​−1)2+2. Choice A is a vertical stretch by factor 3. Choice B represents a horizontal compression by factor 13\frac{1}{3}31​. Choice D incorrectly applies the stretch factor to the entire expression (x−1)(x-1)(x−1) rather than just xxx, which would not preserve the horizontal shift properly.

Question 5

Consider the function m(x)=−12(x+2)3+1m(x) = -\frac{1}{2}(x+2)^3 + 1m(x)=−21​(x+2)3+1. If the point (a,b)(a, b)(a,b) lies on the graph of the parent function f(x)=x3f(x) = x^3f(x)=x3, what point lies on the graph of m(x)m(x)m(x)?

  1. (−12(a+2),b+1)(-\frac{1}{2}(a + 2), b + 1)(−21​(a+2),b+1)
  2. (a+2,−12b+1)(a + 2, -\frac{1}{2}b + 1)(a+2,−21​b+1)
  3. (a−2,12b+1)(a - 2, \frac{1}{2}b + 1)(a−2,21​b+1)
  4. (a−2,−12b+1)(a - 2, -\frac{1}{2}b + 1)(a−2,−21​b+1)
Explanation: When you see a function transformation problem like this, you need to track how each transformation affects the coordinates of points on the original function. Let's analyze m(x)=−12(x+2)3+1m(x) = -\frac{1}{2}(x+2)^3 + 1m(x)=−21​(x+2)3+1 compared to the parent function f(x)=x3f(x) = x^3f(x)=x3. This involves three transformations applied in sequence:
  1. The (x+2)(x+2)(x+2) shifts the graph 2 units left
  2. The −12-\frac{1}{2}−21​ coefficient reflects over the x-axis and compresses vertically by factor 12\frac{1}{2}21​
  3. The +1+1+1 shifts the graph 1 unit up
If point (a,b)(a,b)(a,b) is on f(x)=x3f(x) = x^3f(x)=x3, then b=a3b = a^3b=a3. To find the corresponding point on m(x)m(x)m(x), apply the transformations to the coordinates:
  • Horizontal shift left 2: xxx-coordinate becomes a−2a - 2a−2
  • Vertical reflection and compression: yyy-coordinate becomes −12b-\frac{1}{2}b−21​b
  • Vertical shift up 1: yyy-coordinate becomes −12b+1-\frac{1}{2}b + 1−21​b+1
The transformed point is (a−2,−12b+1)(a-2, -\frac{1}{2}b + 1)(a−2,−21​b+1), which matches choice D. Choice A incorrectly applies the transformations to get the coordinates rather than finding where the point moves. Choice B has the correct yyy-coordinate transformation but wrong xxx-coordinate (a+2a+2a+2 instead of a−2a-2a−2). Choice C misses the reflection, keeping +12b+\frac{1}{2}b+21​b instead of −12b-\frac{1}{2}b−21​b. Strategy tip: For transformation problems, systematically apply each change to the coordinates in order. Remember that f(x+h)f(x+h)f(x+h) shifts left by hhh units, which often trips students up since the sign seems backwards.

Question 6

The function g(x)=−2∣x+3∣−1g(x) = -2|x + 3| - 1g(x)=−2∣x+3∣−1 is a transformation of the parent function f(x)=∣x∣f(x) = |x|f(x)=∣x∣. Which sequence of transformations, applied in order, produces g(x)g(x)g(x) from f(x)f(x)f(x)?

  1. Shift left 3 units, reflect across x-axis, vertical stretch by factor 2, shift down 1 unit
  2. Shift right 3 units, vertical stretch by factor 2, reflect across x-axis, shift down 1 unit
  3. Reflect across x-axis, shift left 3 units, vertical stretch by factor 2, shift down 1 unit
  4. Vertical stretch by factor 2, shift left 3 units, reflect across x-axis, shift up 1 unit
Explanation: To transform f(x)=∣x∣f(x) = |x|f(x)=∣x∣ into g(x)=−2∣x+3∣−1g(x) = -2|x + 3| - 1g(x)=−2∣x+3∣−1: First, shift left 3 units to get ∣x+3∣|x + 3|∣x+3∣. Then reflect across the x-axis and apply vertical stretch by factor 2 to get −2∣x+3∣-2|x + 3|−2∣x+3∣. Finally, shift down 1 unit to get −2∣x+3∣−1-2|x + 3| - 1−2∣x+3∣−1. Choice B incorrectly shifts right instead of left. Choice C applies the reflection before the horizontal shift, which would give a different result. Choice D shifts up instead of down.

Question 7

If g(x)=2x−4−3g(x) = 2\sqrt{x-4} - 3g(x)=2x−4​−3 and f(x)=xf(x) = \sqrt{x}f(x)=x​, which statement about the range of g(x)g(x)g(x) compared to the range of f(x)f(x)f(x) is correct?

  1. The range of g(x)g(x)g(x) is [0,∞)[0, \infty)[0,∞), the same as f(x)f(x)f(x)
  2. The range of g(x)g(x)g(x) is [−3,∞)[-3, \infty)[−3,∞), shifted down 3 units from f(x)f(x)f(x)
  3. The range of g(x)g(x)g(x) is [4,∞)[4, \infty)[4,∞), shifted right 4 units from f(x)f(x)f(x)
  4. The range of g(x)g(x)g(x) is [1,∞)[1, \infty)[1,∞), reflecting the combined transformations
Explanation: The parent function f(x)=xf(x) = \sqrt{x}f(x)=x​ has range [0,∞)[0, \infty)[0,∞). For g(x)=2x−4−3g(x) = 2\sqrt{x-4} - 3g(x)=2x−4​−3: The horizontal shift (x−4x-4x−4) affects the domain but not the range. The vertical stretch by factor 2 would change [0,∞)[0, \infty)[0,∞) to [0,∞)[0, \infty)[0,∞). The vertical shift down 3 units changes the range to [−3,∞)[-3, \infty)[−3,∞). Choice A ignores the vertical shift. Choice C confuses domain and range. Choice D incorrectly calculates the effect of the transformations on the range.

Question 8

The function p(x)=3−x+2+1p(x) = 3^{-x+2} + 1p(x)=3−x+2+1 is a transformation of the parent exponential function f(x)=3xf(x) = 3^xf(x)=3x. Which point on the graph of f(x)f(x)f(x) corresponds to the point (3,2)(3, 2)(3,2) on the graph of p(x)p(x)p(x)?

  1. (1,1)(1, 1)(1,1)
  2. (5,1)(5, 1)(5,1)
  3. (−1,1)(-1, 1)(−1,1)
  4. (1,3)(1, 3)(1,3)
Explanation: We can rewrite p(x)=3−x+2+1=32−x+1p(x) = 3^{-x+2} + 1 = 3^{2-x} + 1p(x)=3−x+2+1=32−x+1. If (3,2)(3, 2)(3,2) is on p(x)p(x)p(x), we verify: p(3)=32−3+1=3−1+1=13+1=43p(3) = 3^{2-3} + 1 = 3^{-1} + 1 = \frac{1}{3} + 1 = \frac{4}{3}p(3)=32−3+1=3−1+1=31​+1=34​. This doesn't equal 2, so let me recalculate. Actually, p(3)=3−3+2+1=3−1+1=13+1=43≠2p(3) = 3^{-3+2} + 1 = 3^{-1} + 1 = \frac{1}{3} + 1 = \frac{4}{3} \neq 2p(3)=3−3+2+1=3−1+1=31​+1=34​=2. Let me check which x gives p(x)=2p(x) = 2p(x)=2: 3−x+2+1=23^{-x+2} + 1 = 23−x+2+1=2, so 3−x+2=1=303^{-x+2} = 1 = 3^03−x+2=1=30, thus −x+2=0-x+2 = 0−x+2=0 and x=2x = 2x=2. So (2,2)(2, 2)(2,2) is on p(x)p(x)p(x). Working backwards from this point: the corresponding point on f(x)=3xf(x) = 3^xf(x)=3x is found by reversing transformations. Since p(x)=3−(x−2)+1p(x) = 3^{-(x-2)} + 1p(x)=3−(x−2)+1, this involves reflection across y-axis and shift up 1. Reversing: (2,2)→(2,1)→(−2,1)(2,2) \rightarrow (2,1) \rightarrow (-2,1)(2,2)→(2,1)→(−2,1) on f(x)f(x)f(x). But this doesn't match our given point. Let me verify the original constraint matches an answer choice by checking f(−1)=3−1=13f(-1) = 3^{-1} = \frac{1}{3}f(−1)=3−1=31​, and working forward through transformations.