Q. 26 - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=\frac{x^2+2}{x+4}$

(a) Is the point $\left(1,\frac{3}{5}\right)$ on the graph of $f$?

(b) If $x=0$, what is $f\left(x\right)$? What point is on the graph of $f$?

(c) If $f\left(x\right)=\frac{1}{2}$, what is $x$? What point(s) are on the graph of $f$?

(d) What is the domain of $f$?

(e) List the x-intercepts, if any, of the graph of $f$.

(f) List the y-intercept, if there is one, of the graph of $f$. 

Step-by-Step Solution

Answer

Part (a) Yes

Part (b) $f\left(0\right)=\frac{1}{2}$; The point on the graph is $\left(0,\frac{1}{2}\right)$.

Part (c) Values of $x$ are $0 and \frac{1}{2}$ ; Points on the graph are $\left(0,\frac{1}{2}\right) and \left(\frac{1}{2},\frac{1}{2}\right)$.

Part (d) The domain is $\left\{x\overline{)x\ne -4}\right\}$.

Part (e) There is no x-intercept.

Part (f) The y-intercept is $\frac{1}{2}$.

Part (a) Step 1. Find the value of the function given point.

When $x=1$,

$\begin{array}{rcl}f\left(x\right)& =& \frac{x^2+2}{x+4}\\ f\left(1\right)& =& \frac{{\left(1\right)}^2+2}{1+4}\\ & =& \frac{1+2}{5}\\ & =& \frac{3}{5}\end{array}$

The point $\left(1,\frac{3}{5}\right)$ is on the graph of $f$.

Part (b) Step 1. Find the value at 2.

If $x=0$, then

$\begin{array}{rcl}f\left(x\right)& =& \frac{x^2+2}{x+4}\\ f\left(0\right)& =& \frac{{\left(0\right)}^2+2}{0+4}\\ & =& \frac{0+2}{4}\\ & =& \frac{1}{2}\end{array}$

The point  $\left(0,\frac{1}{2}\right)$ is on the graph.

Part (c) Step 1. Find the value of x .

If $f\left(x\right)=\frac{1}{2}$, then

$\begin{array}{rcl}f\left(x\right)& =& \frac{1}{2}\\ \frac{x^2+2}{x+4}& =& \frac{1}{2}\\ 2\left(x^2+2\right)& =& x+4\\ 2x^2+4& =& x+4\\ 2x^2-x& =& 0\\ x\left(2x-1\right)& =& 0\\ x& =& 0 or x=\frac{1}{2}\end{array}$

The points on the graph are $\left(0,\frac{1}{2}\right) and \left(\frac{1}{2},\frac{1}{2}\right)$.

Part (d) Step 1. Find the domain.

The domain of $f$ is $\left\{x\overline{)x\ne -4}\right\}$, since $x=-4$ results in division by $0$.  

Part (e) Step 1. Find x -intercepts.

The x-intercepts of the graph of $f$ are the real solutions of the equation $f\left(x\right)=0$ that are in the domain of $f$.

The real solution of the equation $f\left(x\right)=\begin{array}{rcl}& & \frac{x^2+2}{x+4}\end{array}$ is

$\begin{array}{rcl}f\left(x\right)& =& 0\\ \frac{x^2+2}{x+4}& =& 0\\ x^2+2& =& 0\\ x^2& =& -2\end{array}$

There is no real solution for $f\left(x\right)=0$.

There is no x-intercept of the graph $f$.

Part (f) Step 1. Find y -intercepts.

The y-intercepts of the graph of $f$ are the real values of $f\left(0\right)$.

From part (b), $f\left(0\right)=\frac{1}{2}$.

The y-intercept is $\frac{1}{2}$.