College Algebra Quiz: Graphing From Parent Functions

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

1. Horizontal compression by factor $\frac{1}{2}$, shift left 1 unit, shift down 4 units
2. Vertical compression by factor $\frac{1}{2}$, shift left 1 unit, shift down 4 units (correct answer)
3. Vertical compression by factor $\frac{1}{2}$, shift right 1 unit, shift down 4 units
4. Horizontal compression by factor $\frac{1}{2}$, shift right 1 unit, shift up 4 units

Explanation: Starting with $f(x) = x^3$: The term $(x + 1)^3$ represents a horizontal shift left 1 unit. The coefficient $\frac{1}{2}$ represents a vertical compression by factor $\frac{1}{2}$. The $-4$ represents a vertical shift down 4 units. So $g(x) = \frac{1}{2}(x + 1)^3 - 4$. 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

1. $(-\infty, 2]$ (correct answer)
2. $[2, \infty)$
3. $[-2, \infty)$
4. $(-\infty, -2]$

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

Question 3

1. Vertical: $x = 3$, Horizontal: $y = 1$
2. Vertical: $x = -3$, Horizontal: $y = -1$
3. Vertical: $x = 3$, Horizontal: $y = -1$
4. Vertical: $x = -3$, Horizontal: $y = 1$ (correct answer)

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

Question 4

1. $g(x) = 3(x-1)^2 + 2$
2. $g(x) = (3x-1)^2 + 2$
3. $g(x) = \left(\frac{x}{3}-1\right)^2 + 2$ (correct answer)
4. $g(x) = \left(\frac{x-1}{3}\right)^2 + 2$

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

Question 5

1. $(-\frac{1}{2}(a + 2), b + 1)$
2. $(a + 2, -\frac{1}{2}b + 1)$
3. $(a - 2, \frac{1}{2}b + 1)$
4. $(a - 2, -\frac{1}{2}b + 1)$ (correct answer)

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 $m(x) = -\frac{1}{2}(x+2)^3 + 1$ compared to the parent function $f(x) = x^3$. This involves three transformations applied in sequence:

1. The $(x+2)$ shifts the graph 2 units left
2. The $-\frac{1}{2}$ coefficient reflects over the x-axis and compresses vertically by factor $\frac{1}{2}$
3. The $+1$ shifts the graph 1 unit up

• Horizontal shift left 2: $x$-coordinate becomes $a - 2$
• Vertical reflection and compression: $y$-coordinate becomes $-\frac{1}{2}b$
• Vertical shift up 1: $y$-coordinate becomes $-\frac{1}{2}b + 1$

Question 6

1. Shift left 3 units, reflect across x-axis, vertical stretch by factor 2, shift down 1 unit (correct answer)
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 $f(x) = |x|$ into $g(x) = -2|x + 3| - 1$: First, shift left 3 units to get $|x + 3|$. Then reflect across the x-axis and apply vertical stretch by factor 2 to get $-2|x + 3|$. Finally, shift down 1 unit to get $-2|x + 3| - 1$. 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

1. The range of $g(x)$ is $[0, \infty)$, the same as $f(x)$
2. The range of $g(x)$ is $[-3, \infty)$, shifted down 3 units from $f(x)$ (correct answer)
3. The range of $g(x)$ is $[4, \infty)$, shifted right 4 units from $f(x)$
4. The range of $g(x)$ is $[1, \infty)$, reflecting the combined transformations

Explanation: The parent function $f(x) = \sqrt{x}$ has range $[0, \infty)$. For $g(x) = 2\sqrt{x-4} - 3$: The horizontal shift ($x-4$) affects the domain but not the range. The vertical stretch by factor 2 would change $[0, \infty)$ to $[0, \infty)$. The vertical shift down 3 units changes the range to $[-3, \infty)$. 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

1. $(1, 1)$
2. $(5, 1)$
3. $(-1, 1)$ (correct answer)
4. $(1, 3)$

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