Graphing from Parent Functions - College Algebra

1. The function $g(x) = -2|x + 3| - 1$ is a transformation of the parent function $f(x) = |x|$. Which sequence of transformations, applied in order, produces $g(x)$ from $f(x)$?
   Answer: Shift left 3 units, reflect across x-axis, vertical stretch by factor 2, shift down 1 unit 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.
2. If $f(x) = x^3$ and $g(x) = \frac{1}{2}(x + 1)^3 - 4$, then $g(x)$ can be described as $f(x)$ after which transformations?
   Answer: Vertical compression by factor $\frac{1}{2}$, shift left 1 unit, shift down 4 units 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.
3. If $g(x) = 2\sqrt{x-4} - 3$ and $f(x) = \sqrt{x}$, which statement about the range of $g(x)$ compared to the range of $f(x)$ is correct?
   Answer: The range of $g(x)$ is $[-3, \infty)$, shifted down 3 units from $f(x)$ 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.
4. Consider the function $h(x) = \sqrt{4 - 2x} + 3$. This function can be obtained by transforming the parent function $f(x) = \sqrt{x}$. What is the domain of $h(x)$?
   Answer: $(-\infty, 2]$ 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.
5. The function $p(x) = 3^{-x+2} + 1$ is a transformation of the parent exponential function $f(x) = 3^x$. Which point on the graph of $f(x)$ corresponds to the point $(3, 2)$ on the graph of $p(x)$?
   Answer: $(-1, 1)$ 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.
6. Which of the following functions represents a horizontal stretch of $f(x) = (x-1)^2 + 2$ by a factor of 3?
   Answer: $g(x) = \left(\frac{x}{3}-1\right)^2 + 2$ 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.
7. Consider the function $m(x) = -\frac{1}{2}(x+2)^3 + 1$. If the point $(a, b)$ lies on the graph of the parent function $f(x) = x^3$, what point lies on the graph of $m(x)$?
   Answer: $(a - 2, -\frac{1}{2}b + 1)$ 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 If point $(a,b)$ is on $f(x) = x^3$, then $b = a^3$. To find the corresponding point on $m(x)$, apply the transformations to the coordinates: - 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 $-\frac{1}{2}b + 1$ The transformed point is $(a-2, -\frac{1}{2}b + 1)$, 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 $y$-coordinate transformation but wrong $x$-coordinate ($a+2$ instead of $a-2$). Choice C misses the reflection, keeping $+\frac{1}{2}b$ instead of $-\frac{1}{2}b$. Strategy tip: For transformation problems, systematically apply each change to the coordinates in order. Remember that $f(x+h)$ shifts left by $h$ units, which often trips students up since the sign seems backwards.
8. The parent function $f(x) = \frac{1}{x}$ is transformed to create $h(x) = \frac{-2}{x+3} + 1$. What are the equations of the vertical and horizontal asymptotes of $h(x)$?
   Answer: Vertical: $x = -3$, Horizontal: $y = 1$ 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 $+3$.

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