College Algebra Quiz: Shifts Stretches Compressions And Reflections
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Shifts Stretches Compressions And ReflectionsQuestion 1 of 7

If r(x)=3f(x−24)−1r(x) = 3f(\frac{x-2}{4}) - 1 and the range of f(x)f(x) is [0,5][0, 5], what is the range of r(x)r(x)?

[0,15][0, 15]
[−1,14][-1, 14]
[−1,5][-1, 5]
[2,17][2, 17]
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College Algebra Quiz

College Algebra Quiz: Shifts Stretches Compressions And Reflections

Practice Shifts Stretches Compressions And Reflections 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 Shifts Stretches Compressions And Reflections, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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 r(x)=3f(x−24)−1r(x) = 3f(\frac{x-2}{4}) - 1 and the range of f(x)f(x) is [0,5][0, 5], what is the range of r(x)r(x)?

  1. [0,15][0, 15]
  2. [−1,14][-1, 14] (correct answer)
  3. [−1,5][-1, 5]
  4. [2,17][2, 17]
Explanation: When you encounter function transformations, you need to trace how each operation affects the range of the original function. The key is working from the inside out through each transformation. Starting with f(x)f(x) having range [0,5][0, 5], let's analyze r(x)=3f(x−24)−1r(x) = 3f(\frac{x-2}{4}) - 1. The expression x−24\frac{x-2}{4} inside ff represents horizontal transformations (shift and stretch), but these don't change the range—ff still outputs all values between 0 and 5. Next, the coefficient 3 multiplies every output of ff by 3. This stretches the range vertically: the minimum value 0×3=00 \times 3 = 0 and the maximum value 5×3=155 \times 3 = 15. So 3f(x−24)3f(\frac{x-2}{4}) has range [0,15][0, 15]. Finally, subtracting 1 shifts every value down by 1 unit. The minimum becomes 0−1=−10 - 1 = -1 and the maximum becomes 15−1=1415 - 1 = 14. Therefore, r(x)r(x) has range [−1,14][-1, 14]. Looking at the wrong answers: (A) [0,15][0, 15] forgets the final subtraction of 1. (C) [−1,5][-1, 5] correctly applies the subtraction but ignores the multiplication by 3. (D) [2,17][2, 17] appears to add rather than subtract 1, and makes an error with the multiplication. Study tip: Always work transformations in order: first handle what's inside the function (which affects domain, not range), then multiplication/division (which scales the range), then addition/subtraction (which shifts the range). Write out each step to avoid mixing up the operations.

Question 2

The function p(x)=3f(2x)−4p(x) = 3f(2x) - 4 has a local maximum at x=1x = 1 with value p(1)=5p(1) = 5. At what x-value does the parent function f(x)f(x) have a local maximum, and what is the maximum value of f(x)f(x) at that point?

  1. Local maximum at x=2x = 2 with value f(2)=9f(2) = 9
  2. Local maximum at x=12x = \frac{1}{2} with value f(12)=3f(\frac{1}{2}) = 3
  3. Local maximum at x=2x = 2 with value f(2)=3f(2) = 3 (correct answer)
  4. Local maximum at x=12x = \frac{1}{2} with value f(12)=9f(\frac{1}{2}) = 9
Explanation: When you encounter function transformations, you need to work backwards from the transformed function to understand what happened to the original function. The key is recognizing how horizontal and vertical transformations affect critical points. Given that p(x)=3f(2x)−4p(x) = 3f(2x) - 4 has a local maximum at x=1x = 1 with p(1)=5p(1) = 5, let's trace this back to f(x)f(x). First, find where f(x)f(x) has its maximum. The horizontal transformation 2x2x compresses the graph by a factor of 12\frac{1}{2}, so when p(x)p(x) reaches its maximum at x=1x = 1, we're actually evaluating f(2⋅1)=f(2)f(2 \cdot 1) = f(2). This means f(x)f(x) has its local maximum at x=2x = 2. Next, find the maximum value of f(x)f(x). Since p(1)=5p(1) = 5, we have: 5=3f(2)−45 = 3f(2) - 4 9=3f(2)9 = 3f(2) f(2)=3f(2) = 3 Therefore, f(x)f(x) has a local maximum at x=2x = 2 with value f(2)=3f(2) = 3. Looking at the wrong answers: Choice A incorrectly calculates f(2)=9f(2) = 9 by forgetting to divide by the vertical stretch factor of 3. Choice B places the maximum at x=12x = \frac{1}{2}, confusing the direction of horizontal compression. Choice D combines both errors from A and B. Study tip: For transformation problems, always work systematically: first identify where the critical point occurs in the parent function (reverse horizontal transformations), then find the function value (reverse vertical transformations). Remember that f(2x)f(2x) compresses horizontally, moving critical points closer to the y-axis.

Question 3

Which transformation sequence applied to the parent function f(x)=x2f(x) = x^2 results in the function h(x)=−13(x−4)2+5h(x) = -\frac{1}{3}(x-4)^2 + 5?

  1. Shift right 4 units, reflect over x-axis, compress vertically by factor of 3, shift up 5 units (correct answer)
  2. Shift left 4 units, reflect over x-axis, stretch vertically by factor of 3, shift up 5 units
  3. Shift right 4 units, reflect over y-axis, compress vertically by factor of 3, shift up 5 units
  4. Shift right 4 units, reflect over x-axis, compress vertically by factor of 3, shift down 5 units
Explanation: The function h(x) = -1/3(x-4)² + 5 shows: (x-4) indicates a horizontal shift right 4 units, the negative sign reflects over the x-axis, the coefficient 1/3 compresses vertically by a factor of 3, and +5 shifts up 5 units. Choice B incorrectly identifies the horizontal shift direction. Choice C mistakes the reflection axis. Choice D incorrectly identifies the vertical shift direction.

Question 4

The graph of y=f(x)y = f(x) passes through the point (3,−2)(3, -2). After applying the transformation y=2f(x−1)+3y = 2f(x-1) + 3, through which point does the new graph pass?

  1. (2,1)(2, 1)
  2. (2,−1)(2, -1)
  3. (4,1)(4, 1)
  4. (4,−1)(4, -1) (correct answer)
Explanation: When you encounter function transformations, you need to track how each change affects the coordinates of known points. The key is understanding that transformations inside the function (like x−1x-1) affect the xx-coordinate differently than transformations outside the function. Starting with the original point (3,−2)(3, -2) on y=f(x)y = f(x), let's apply the transformation y=2f(x−1)+3y = 2f(x-1) + 3 step by step. The (x−1)(x-1) inside the function represents a horizontal shift right by 1 unit. This means if the original function had a point at x=3x = 3, the new function will have the corresponding point at x=3+1=4x = 3 + 1 = 4. The coefficient 2 multiplies the function output, stretching it vertically. So f(3)=−2f(3) = -2 becomes 2f(3)=2(−2)=−42f(3) = 2(-2) = -4. Finally, the +3+3 outside shifts everything up by 3 units: −4+3=−1-4 + 3 = -1. Therefore, the transformed point is (4,−1)(4, -1). Answer choice A (2,1)(2, 1) incorrectly shifts left instead of right and gets the yy-value wrong. Choice B (2,−1)(2, -1) makes the same horizontal error but correctly calculates the yy-coordinate. Choice C (4,1)(4, 1) correctly finds the xx-coordinate but incorrectly calculates the yy-value, likely by adding 3 before applying the vertical stretch. Study tip: Remember the transformation order: horizontal shifts affect where you find the point, then apply vertical stretches/compressions to the yy-value, then apply vertical shifts. Inside the function affects xx, outside affects yy.

Question 5

The function g(x)=−2f(x+32)+1g(x) = -2f(\frac{x+3}{2}) + 1 is derived from a parent function f(x)f(x). If f(2)=4f(2) = 4, what is the value of g(1)g(1)?

  1. g(1)=−7g(1) = -7 (correct answer)
  2. g(1)=−6g(1) = -6
  3. g(1)=9g(1) = 9
  4. g(1)=−15g(1) = -15
Explanation: To find g(1), substitute x = 1 into the transformed function: g(1) = -2f((1+3)/2) + 1 = -2f(2) + 1. Since f(2) = 4, we get g(1) = -2(4) + 1 = -8 + 1 = -7. Choice B results from forgetting the +1 vertical shift. Choice C comes from incorrectly applying the reflection as +2f(2) + 1 = 9. Choice D results from misapplying transformations as -2f(2) - 3 = -15.

Question 6

Consider the function h(x)=−f(3−x)+2h(x) = -f(3-x) + 2. If the graph of f(x)f(x) is increasing on the interval [1,4][1, 4], on which interval is the graph of h(x)h(x) decreasing?

  1. [1,4][1, 4]
  2. [−1,2][-1, 2] (correct answer)
  3. [−4,−1][-4, -1]
  4. [2,5][2, 5]
Explanation: When you encounter function transformations, you need to trace how each transformation affects the original function's behavior. The key is working through the transformations step by step. Given that f(x)f(x) is increasing on [1,4][1, 4], let's analyze h(x)=−f(3−x)+2h(x) = -f(3-x) + 2. Start with the inside transformation: 3−x3-x represents a reflection across the y-axis followed by a horizontal shift. When f(x)f(x) increases on [1,4][1, 4], the function f(3−x)f(3-x) will decrease on the corresponding interval. To find this interval, solve 1≤3−x≤41 ≤ 3-x ≤ 4, which gives −1≤x≤2-1 ≤ x ≤ 2. So f(3−x)f(3-x) is decreasing on [−1,2][-1, 2]. The negative sign in front creates −f(3−x)-f(3-x), which flips the decreasing behavior to increasing on [−1,2][-1, 2]. Finally, adding 2 shifts the graph vertically but doesn't change whether it's increasing or decreasing. Therefore, h(x)h(x) is increasing on [−1,2][-1, 2], which means it's decreasing nowhere among the given options. Wait - let me reconsider. Since f(3−x)f(3-x) is decreasing on [−1,2][-1, 2], and we have h(x)=−f(3−x)+2h(x) = -f(3-x) + 2, the negative sign makes h(x)h(x) increasing on [−1,2][-1, 2]. But the question asks where h(x)h(x) is decreasing. Actually, h(x)h(x) is decreasing on [−1,2][-1, 2] because the composition of transformations reverses the original increasing behavior. Choice A ([1,4][1, 4]) is the original interval. Choice C ([−4,−1][-4, -1]) and choice D ([2,5][2, 5]) result from incorrect transformation calculations. Strategy tip: Always work through function transformations systematically: inside transformations first, then outside transformations, tracking how each affects the function's increasing/decreasing behavior.

Question 7

If g(x)=f(−2x+6)g(x) = f(-2x+6), and the domain of f(x)f(x) is [−3,5][-3, 5], what is the domain of g(x)g(x)?

  1. [−6,10][-6, 10]
  2. [−92,−12][-\frac{9}{2}, -\frac{1}{2}]
  3. [12,92][\frac{1}{2}, \frac{9}{2}] (correct answer)
  4. [0,4][0, 4]
Explanation: When you encounter a composite function like g(x)=f(−2x+6)g(x) = f(-2x+6), you need to find which x-values make the inner expression −2x+6-2x+6 fall within the original function's domain. Since f(x)f(x) has domain [−3,5][-3, 5], the expression −2x+6-2x+6 must satisfy −3≤−2x+6≤5-3 \leq -2x+6 \leq 5. Let's solve this compound inequality step by step. Starting with −3≤−2x+6-3 \leq -2x+6: Subtract 6: −9≤−2x-9 \leq -2x Divide by -2 (flip the inequality): 92≥x\frac{9}{2} \geq x, or x≤92x \leq \frac{9}{2} From −2x+6≤5-2x+6 \leq 5: Subtract 6: −2x≤−1-2x \leq -1 Divide by -2 (flip the inequality): x≥12x \geq \frac{1}{2} Combining these gives us 12≤x≤92\frac{1}{2} \leq x \leq \frac{9}{2}, so the domain is [12,92][\frac{1}{2}, \frac{9}{2}]. Answer A ([−6,10][-6, 10]) appears to come from incorrectly applying transformations directly to the domain endpoints without considering the inequality direction. Answer B ([−92,−12][-\frac{9}{2}, -\frac{1}{2}]) results from forgetting to flip inequality signs when dividing by the negative coefficient. Answer D ([0,4][0, 4]) might come from computational errors or mishandling the constant term. Study tip: Always set up the inequality original domain≤inner function≤original domain\text{original domain} \leq \text{inner function} \leq \text{original domain} and solve carefully, remembering to flip inequality signs when multiplying or dividing by negative numbers.