Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

Suppose that the $x$-intercepts of the graph of $y=f\left(x\right)$ are$-8$ and $1$.

(a) What are the x-intercepts of the graph of $y=f(x+4)$?

(b) What are the x-intercepts of the graph of $y=f(x-3)$?

(c) What are the x-intercepts of the graph of $y=2f\left(x\right)$?

(d) What are the $x$-intercepts of the graph of $y=f(-x)$? 

Step-by-Step Solution

Answer

Part a. The x intercepts of $y=f(x+4)$ are $-12$ and $-3$.

Part b. The x intercepts of $y=f(x-3)$ are $-5$ and $4$.

Part c. The x intercepts of $y=2f\left(x\right)$ are $-8$ and $1$.

Part d. The x intercepts of $y=f(-x)$ are $8$ and $1$.

Part (a) Step 1. Find the x intercepts

The graph of $y=f(x+4)$is obtained by horizontally shifting the graph of $y=f\left(x\right)$ four units left.

So if the x intercepts of $y=f\left(x\right)$ are $-8$ and $1$.

Then the x intercepts of $y=f(x+4)$ are given as

$\begin{array}{rcl}x+4& =& -8\\ x+4-4& =& -8-4\\ x& =& -12\end{array}$

and

$\begin{array}{rcl}x+4& =& 1\\ x+4-4& =& 1-4\\ x& =& -3\end{array}$

Part (b) Step 1. Find the x intercepts

The graph of $y=f(x-3)$ is obtained by horizontally shifting the graph of $y=f\left(x\right)$ three units right.

So if the x intercepts of $y=f\left(x\right)$ are $-8$ and $1$.

Then the x intercepts of $y=f(x-3)$ are given as

$\begin{array}{rcl}x-3& =& -8\\ x-3+3& =& -8+3\\ x& =& -5\end{array}$

and

$\begin{array}{rcl}x-3& =& 1\\ x-3+3& =& 1+3\\ x& =& 4\end{array}$

Part (c) Step 1. Find the x intercepts

The function $y=2f\left(x\right)$ is zero when $y=f\left(x\right)$ is zero.

So the two functions have the same $x$ intercepts.

So the x intercepts of $y=2f\left(x\right)$ are $-8$ and $1$.

Part (d) Step 1. Find the x intercepts

The graph of $y=f(-x)$ is a reflection of the graph $y=f\left(x\right)$ about the $y$ axis.

So if the x intercepts of $y=f\left(x\right)$ are $-8$ and $1$.

Then the x intercepts of $y=f(-x)$ is given as

$\begin{array}{rcl}-x& =& -8\\ -x·(-1)& =& -8·(-1)\\ x& =& 8\end{array}$

and

$\begin{array}{rcl}-x& =& 1\\ -x·(-1)& =& 1·(-1)\\ x& =& -1\end{array}$