Chapter 2.2, Problem 26AYU

In Problems 25-30, answer the questions about the given function.

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

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

(b) If $x=\text{ 0}$. 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 $x\text{-intercepts}$, if any, of the graph of $f$.

(f) List the $y\text{-intercepts}$, if there is one, of the graph of $f$.

To determine

To find: When $f(x)=\frac{x^2+2}{x+4}$, answer the following

a. Is the point $(1,\frac{3}{5})$ 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 of $x\text{-intercepts}$, if any, of the graph of $f$.

f. List of the $y\text{-intercept}$, if there is one, of the graph of $f$.

Answer to Problem 26AYU

Solution:

a. Yes, the point $(1,\frac{3}{5})$ is on the graph of $f$.

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

c. 0, $\frac{1}{2}$; The points $(0, \frac{1}{2})$ and $(\frac{1}{2},\frac{1}{2})$ are on the graph of $f$.

d. Domain of $f=\left\{x|x\ne -4\right\}$.

e. No $x\text{-intercept}$.

f. $y\text{-intercept}=\frac{1}{2}$.

Explanation of Solution

Given:

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

Calculation:

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

To find the point $(1,\frac{3}{5})$ on the graph of $f$, substitute $x=1$ in $f(x)$.

$f(1)=\frac{1^2+2}{1+4}$

$=\frac{3}{5}$

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

b. If $x=0$.

$f(0)=\frac{0^2+2}{0+4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

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

c. $f(x)=\frac{1}{2}\Rightarrow \frac{x^2+2}{x+4}=\frac{1}{2}$

$2\left(x²+2\right)=1\left(x+4\right)$

$2x²+4=x+4$

$2x²-x=0$

$x\left(2x-1\right)=0$

$x=0;2x-1=0$

$x=0;x=\frac{1}{2}$

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

d. Domain of $f=\left\{x|x\ne -4\right\}$.

e. To find $x\text{-intercept}$, let $f\left(x\right)=0$,

$\frac{x^2+2}{x+4}=0$

$x²+2=0$

$x²=-2 \Rightarrow x$ is not real

f. To find the $y\text{-intercept}$, let $x=0$ in $f(x)=\frac{x^2+2}{x+4}$.

$f(0)=\frac{0^2+2}{0+4}=\frac{2}{4}=\frac{1}{2}\Rightarrow y\text{-intercept}=\frac{1}{2}$