Transformations of Functions
Help Questions
AP Precalculus › Transformations of Functions
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 $$h(x)$$?
Domain: all real numbers except $$x=-2$$. Range: all real numbers except $$y=6$$.
Domain: all real numbers except $$x=6$$. Range: all real numbers except $$y=6$$.
Domain: all real numbers except $$x=-2$$. Range: all real numbers except $$y=-4$$.
Domain: all real numbers except $$x=6$$. Range: all real numbers except $$y=-4$$.
Explanation
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 $$y=1-5=-4$$.
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 $$g(x)$$?
As $$x \to \infty$$, $$g(x) \to \infty$$, and as $$x \to -\infty$$, $$g(x) \to \infty$$.
As $$x \to \infty$$, $$g(x) \to -\infty$$, and as $$x \to -\infty$$, $$g(x) \to \infty$$.
As $$x \to \infty$$, $$g(x) \to -\infty$$, and as $$x \to -\infty$$, $$g(x) \to -\infty$$.
As $$x \to \infty$$, $$g(x) \to \infty$$, and as $$x \to -\infty$$, $$g(x) \to -\infty$$.
Explanation
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 $$p(x)$$
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 $$g(x)$$?
$$g(x) = \frac{1}{-3x} + 1$$
$$g(x) = -\frac{1}{3x} + 1$$
$$g(x) = \frac{-1}{\frac{1}{3}x} + 1$$
$$g(x) = \frac{1}{-\frac{1}{3}x} + 1$$
Explanation
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 $$f(x)$$?
The vertical asymptote shifts from $$x=0$$ to $$x=1$$, and the horizontal asymptote shifts from $$y=0$$ to $$y=3$$.
The vertical asymptote shifts from $$x=0$$ to $$x=-1$$, and the horizontal asymptote shifts from $$y=0$$ to $$y=3$$.
The vertical asymptote is stretched from $$x=0$$ to $$x=2$$, and the horizontal asymptote is shifted from $$y=0$$ to $$y=3$$.
The vertical asymptote shifts from $$x=0$$ to $$x=1$$, and the horizontal asymptote is stretched from $$y=0$$ to $$y=2$$.
Explanation
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 $$y=3$$.