Question

Plot points between and beyond each X intercept and vertical asymptote find the value of the function at the given value of XFractions please

Answer 1

Given the function:

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

when x=-2, f(x) = ?

Substitute the value of -2 for x in the f(x) function.

$\begin{gathered} f(x)=\frac{2x^2}{x^2+4} \\ f(-2)=\frac{2(-2)^2}{(-2)^2+4}=1 \end{gathered}$

when x = -1, f(x) =?

Substitute the value of -1 for x in the f(x) function.

$\begin{gathered} f(x)=\frac{2x^2}{x^2+4} \\ f(-1)=\frac{2(-1)^2}{(-1)^2+4}=\frac{2}{5} \end{gathered}$

when x = 5, f(x) =?

Substitute the value of 5 for x in the f(x) function.

$\begin{gathered} f(x)=\frac{2x^2}{x^2+4} \\ f(5)=\frac{2(5)^2}{(5)^2+4}=\frac{50}{29} \end{gathered}$

when x = 6, f(x) =?

Substitute the value of 6 for x in the f(x) function.

$\begin{gathered} f(x)=\frac{2x^2}{x^2+4} \\ f(6)=\frac{2(6)^2}{(6)^2+4}=\frac{9}{5} \end{gathered}$

Thus, we have

$\begin{gathered} x=-2,\text{ f(}x)\text{ =}1 \\ x=-1,\text{ f(x)=}\frac{2}{5} \\ x=5,\text{ f(x)=}\frac{50}{29} \\ x=6,\text{ f(x)=}\frac{9}{5} \end{gathered}$

The graph of the f(x) function is as shown below: