College Algebra, Chapter 4, 4.2, Section 4.2, Problem 8

Sketch the graph of each function by applying sketch transformations on the standard form $y = x^n$. Indicate all $x$ and $y$-intercept on each graph.
a.) $P(x) = (x + 3)^5$
b.) $Q(x) = -2(x + 3)^5 - 64$
c.) $\displaystyle R(x) = -\frac{1}{2} (x - 2)^5$
d.) $\displaystyle S(x) = -\frac{1}{2} (x - 2)^5 + 16$

a.) The graph of $P(x)$ is the graph of $y = x^5$ that is shifted 3 units to the left. Its $y$-intercept is 243 and $x$-intercept is $-3$.



b.) The graph of $Q(x)$ is the graph of $y = x^5$ that is reflected about $x$-axis, then stretched vertically by a factor of 2. Lastly, the result is shifted 64 units downward. To solve for the $y$-intercept, we set $x = 0$
$y = -2(0 + 3)^5 - 64 = -550$

To solve for the $x$-intercept, we set $y = 0$

$\begin{equation} \begin{aligned} 0 &= -2 ( x + 3 )^5 - 64\\ \\ -2 ( x + 3 )^5 &= 64\\ \\ ( x + 3 )^5 &= -32\\ \\ x + 3 &= -2 \\ \\ x &= -5 \end{aligned} \end{equation}$




c.) The graph of $R(x)$ is the graph of $y = x^5$ that is shifted 2 units to the right and reflected about $x$-axis. Lastly, the result is compressed vertically by a factor of 2. To solve for $y$-intercepts, we set $x = 0$
$\displaystyle y = -\frac{1}{2} ( 0 - 2 )^5 = -\frac{1}{2} (-32) = 16$
To solve for $x$-intercept, we set $y = 0$

$\begin{equation} \begin{aligned} 0 &= -\frac{1}{2} (x - 2)^5\\ \\ (x-2)^5 &= 0 \\ \\ x &= 2 \end{aligned} \end{equation}$




d.) The graph $S(x)$ is the graph of $y = x^5$ that is shifted 2 units to the right, reflected about $x$-axis and compressed vertically by a factor of 2. Lastly, the result is shifted 16 units upward. To solve for $y$-intercept, we set $x = 0$

$\begin{equation} \begin{aligned} y &= -\frac{1}{2} ( 0 - 2 )^5 + 16\\ \\ y &= 16 + 16 = 32 \end{aligned} \end{equation}$

To solve for $x$-intercept, we set $y = 0$

$\begin{equation} \begin{aligned} 0 &= -\frac{1}{2} ( x - 2)^5 + 16\\ \\ -\frac{1}{2} (x - 2)^5 &= -16\\ \\ (x - 2)^5 &= 32\\ \\ x - 2 &= 2\\ \\ x &= 4 \end{aligned} \end{equation}$