Transformations of Functions - AP Precalculus

The graph of $y = x^3$ is transformed to produce the graph of $y = -(x+4)^3$ . Which of the following describes this transformation?
Answer: A reflection across the x-axis, followed by a shift 4 units to the left. The term $(x+4)$ inside the function indicates a horizontal shift of 4 units to the left. The negative sign outside the function indicates a reflection across the x-axis. The order of these two transformations does not matter in this case. Note that for an odd function like $f(x)=x^3$ , a reflection across the y-axis, $f(-x)=(-x)^3=-x^3$ , is equivalent to a reflection across the x-axis, $-f(x) = -x^3$ . However, the shift is unambiguously to the left.
Let $f(x)$ be a polynomial function. The graph of $g(x) = f(x) + c$ , where $c$ is a constant, is a vertical translation of the graph of $f(x)$ . How does the average rate of change of $g(x)$ over the interval $[a, b]$ compare to the average rate of change of $f(x)$ over the same interval?
Answer: The average rate of change of $g(x)$ is the same as the average rate of change of $f(x)$ . The average rate of change of $g(x)$ over $[a, b]$ is $\frac{g(b)-g(a)}{b-a}$ . Substituting $g(x) = f(x)+c$ , this becomes $\frac{(f(b)+c) - (f(a)+c)}{b-a} = \frac{f(b)-f(a)}{b-a}$ , which is the average rate of change of $f(x)$ over $[a, b]$ . A vertical shift does not affect the average rate of change.
The domain of the rational function $r(x)$ is all real numbers except $x=2$ , and its range is all real numbers except $y=1$ . Let $h(x) = r(x+4) - 5$ . What are the domain and range of $h(x)$ ?
Answer: Domain: all real numbers except $x=-2$ . Range: all real numbers except $y=-4$ . The transformation $r(x+4)$ shifts the graph of $r(x)$ four units to the left. This shifts the vertical asymptote and the excluded value in the domain from $x=2$ to $x=2-4=-2$ . The transformation $-5$ shifts the graph five units down. This shifts the horizontal asymptote and the excluded value in the range from $y=1$ to $y=1-5=-4$ .
The function $f(x) = \frac{1}{x}$ is transformed into the function $g(x)$ by a reflection across the y-axis, a horizontal stretch by a factor of 3, and a vertical shift up by 1 unit. Which of the following is an equation for $g(x)$ ?
Answer: $g(x) = \frac{1}{-\frac{1}{3}x} + 1$ A reflection across the y-axis means replacing $x$ with $-x$ . A horizontal stretch by a factor of 3 means replacing $x$ with $(\frac{1}{3}x)$ . Combining these gives replacing $x$ with $(-\frac{1}{3}x)$ . A vertical shift up by 1 unit means adding 1 to the function. So, $g(x) = f(-\frac{1}{3}x) + 1 = \frac{1}{-\frac{1}{3}x} + 1$ .
The function $g(x) = \frac{2}{x-1} + 3$ is obtained by transforming the parent function $f(x) = \frac{1}{x}$ . Which statement accurately describes the effect of these transformations on the asymptotes of $f(x)$ ?
Answer: The vertical asymptote shifts from $x=0$ to $x=1$ , and the horizontal asymptote shifts from $y=0$ to $y=3$ . The parent function $f(x) = 1/x$ has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=0$ . The function $g(x)$ can be written as $g(x) = 2f(x-1) + 3$ . The horizontal shift of 1 unit right changes the vertical asymptote to $x=1$ . The vertical stretch by a factor of 2 does not move the horizontal asymptote from $y=0$ , but the vertical shift of 3 units up moves it to $y=3$ .
The graph of the function $f(x) = x^4$ is transformed to obtain the graph of $g(x) = (2x+8)^4 - 1$ . Which of the following sequences of transformations maps the graph of $f$ to the graph of $g$ ?
Answer: A horizontal compression by a factor of $1/2$ , a horizontal shift 4 units to the left, and a vertical shift 1 unit down. To identify the horizontal transformations correctly, the expression inside the function must be factored: $g(x) = (2(x+4))^4 - 1$ . This shows a horizontal compression by a factor of $1/2$ (from the factor of 2) and a horizontal shift 4 units to the left (from the $(x+4)$ term). The $-1$ outside the function represents a vertical shift 1 unit down.
Let $f(x)$ be a non-constant polynomial function. Which of the following transformations, when applied to the graph of $f(x)$ to produce a new function $g(x)$ , results in $g(x)$ having the same set of real zeros as $f(x)$ ?
Answer: $g(x) = 2f(x)$ A real zero of a function occurs at an x-value where the function's output is 0. If $c$ is a zero of $f(x)$ , then $f(c) = 0$ . For the transformation $g(x) = 2f(x)$ , the new output at $x=c$ is $g(c) = 2f(c) = 2(0) = 0$ . Therefore, $c$ is also a zero of $g(x)$ . Horizontal shifts and compressions change the x-values of the zeros. A vertical shift changes the output values, so if $f(c)=0$ , then $f(c)+1 = 1$ , meaning $c$ is no longer a zero.
The cost, in dollars, to manufacture $x$ widgets is given by the polynomial function $C(x) = 0.1x^3 - 2x^2 + 50x + 200$ . Due to a new supplier, the cost for each widget is reduced, which can be modeled by a horizontal stretch of the graph of $C(x)$ by a factor of 2. Additionally, a new factory fee adds a fixed $_$ 50 to the total cost. Which function $N(x)$ represents the new cost?
Answer: $N(x) = C(\frac{1}{2}x) + 50$ A horizontal stretch by a factor of 2 corresponds to replacing $x$ with $(\frac{1}{2}x)$ in the function definition. This transformation is applied first. A fixed fee of $_$ 50 added to the total cost corresponds to a vertical shift up by 50 units, which means adding 50 to the entire function. Therefore, the new cost function is $N(x) = C(\frac{1}{2}x) + 50$ .
The graph of $y=f(x)$ is transformed to the graph of $y=f(3x-6)$ . This results in which of the following transformations?
Answer: A horizontal compression by a factor of $1/3$ followed by a horizontal shift of 2 units to the right. To correctly identify the transformations, the expression inside the function must be factored: $3x-6 = 3(x-2)$ . The factor of 3 corresponds to a horizontal compression by a factor of $1/3$ . The term $(x-2)$ corresponds to a horizontal shift of 2 units to the right. The standard order is to apply stretches/compressions/reflections first, then shifts.
Let $p(x)$ be a polynomial function of odd degree with a positive leading coefficient. Let $g(x) = -p(-x)$ . Which of the following describes the end behavior of $g(x)$ ?
Answer: As $x \to \infty$ , $g(x) \to \infty$ , and as $x \to -\infty$ , $g(x) \to -\infty$ . A polynomial $p(x)$ of odd degree with a positive leading coefficient has the end behavior: as $x \to \infty$ , $p(x) \to \infty$ , and as $x \to -\infty$ , $p(x) \to -\infty$ . The function $g(x) = -p(-x)$ is a reflection of $p(x)$ across the y-axis, followed by a reflection across the x-axis. To find the end behavior of $g(x)$ , we consider the limits. As $x \to \infty$ , the input to $p$ is $-x$ , which approaches $-\infty$ . So, $p(-x) \to -\infty$ . Then $g(x) = -p(-x) \to -(-\infty) = \infty$ . As $x \to -\infty$ , the input to $p$ is $-x$ , which approaches $\infty$ . So, $p(-x) \to \infty$ . Then $g(x) = -p(-x) \to -(\infty) = -\infty$ . Thus, the end behavior of $g(x)$ is the same as $p(x)$
Let $p(x) = x^2 - 6x$ . The graph of the function $q(x)$ is obtained by vertically stretching the graph of $p(x)$ by a factor of 3 and then reflecting it across the x-axis. Which of the following is the equation for $q(x)$ ?
Answer: $q(x) = -3x^2 + 18x$ A vertical stretch by a factor of 3 means multiplying the function by 3, resulting in $3p(x) = 3(x^2 - 6x) = 3x^2 - 18x$ . A reflection across the x-axis means multiplying the result by -1. Therefore, $q(x) = -1(3p(x)) = -(3x^2 - 18x) = -3x^2 + 18x$ .
The function $f$ is transformed to the function $g$ where $g(x) = -f(x+3)$ . If the point $(-1, 4)$ lies on the graph of $f$ , which of the following points must lie on the graph of $g$ ?
Answer: $(-4, -4)$ The point $(-1, 4)$ on the graph of $f$ means $f(-1) = 4$ . The transformation involves a horizontal shift and a vertical reflection. The horizontal shift is 3 units to the left. The new x-coordinate satisfies $x_{new} + 3 = -1$ , so $x_{new} = -4$ . The vertical transformation is a reflection across the x-axis. The new y-coordinate is $-f(-1) = -4$ . Therefore, the point $(-4, -4)$ must be on the graph of $g$ .
The function $g(x)$ is a transformation of a parent rational function $f(x)$ . If $g(x) = f(-2x)$ , which statement describes the transformations applied to the graph of $f(x)$ to obtain the graph of $g(x)$ ?
Answer: A horizontal compression by a factor of $\frac{1}{2}$ and a reflection across the y-axis. The transformation is of the form $g(x) = f(bx)$ where $b = -2$ . The negative sign indicates a reflection across the y-axis. The magnitude $|b| = 2$ indicates a horizontal compression by a factor of $\frac{1}{|b|} = \frac{1}{2}$ .
Let $f(x) = x^3 - 4x + 1$ . The function $g(x)$ is the result of shifting the graph of $f(x)$ two units to the left and five units up. Which of the following defines the function $g(x)$ ?
Answer: $g(x) = (x+2)^3 - 4(x+2) + 6$ To shift the graph of $f(x)$ two units to the left, we replace $x$ with $(x+2)$ . To shift the graph five units up, we add 5 to the function. Thus, $g(x) = f(x+2) + 5 = ((x+2)^3 - 4(x+2) + 1) + 5 = (x+2)^3 - 4(x+2) + 6$ .
The point $(3, -2)$ is on the graph of the function $y = f(x)$ . Which point must be on the graph of the function $y = 2f(x-1)$ ?
Answer: $(4, -4)$ The transformation $y = 2f(x-1)$ involves a horizontal shift and a vertical stretch. The horizontal transformation is $x \to x-1$ . To find the new x-coordinate, we set the new input equal to the old input: $x_{new}-1 = 3$ , so $x_{new} = 4$ . The vertical transformation is $y \to 2y$ . The new y-coordinate is $2 \times (-2) = -4$ . Thus, the point $(3, -2)$ on the graph of $f(x)$ corresponds to the point $(4, -4)$ on the graph of $y = 2f(x-1)$
The graph of the rational function $g(x) = \frac{1}{x-3} + 5$ is a transformation of the graph of the parent function $f(x) = \frac{1}{x}$ . Which of the following statements correctly describes the transformations?
Answer: A translation 3 units to the right and 5 units up. The function $g(x) = f(x-3) + 5$ . The term $(x-3)$ indicates a horizontal translation of 3 units to the right. The term $+5$ outside the function indicates a vertical translation of 5 units up.
The rational function $h(x)$ has a single vertical asymptote at $x=5$ . The function $k(x)$ is defined by $k(x) = h(2x) - 3$ . What is the vertical asymptote of the graph of $k(x)$ ?
Answer: $x = 2.5$ The vertical asymptote of $h(x)$ occurs where its input is 5. For the function $k(x)$ , the input to $h$ is $2x$ . Therefore, the vertical asymptote of $k(x)$ will occur when $2x = 5$ . Solving for $x$ gives $x = 5/2 = 2.5$ . The vertical shift of $-3$ does not affect the location of the vertical asymptote.
The graph of the rational function $f(x) = \frac{2x^2+1}{x^2-4}$ has a horizontal asymptote at $y=2$ . If the function $g(x) = -f(x) + 3$ , what is the horizontal asymptote of the graph of $g(x)$ ?
Answer: $y = 1$ The transformation $-f(x)$ reflects the graph of $f(x)$ across the x-axis. This transformation reflects the horizontal asymptote from $y=2$ to $y=-2$ . The transformation $+3$ then shifts the graph, including the asymptote, up by 3 units. The new horizontal asymptote is at $y = -2 + 3 = 1$ .

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