91Ó°ÊÓ

The function is given by,

$f\left(x\right)=\sqrt{16+4x^2}$

To graph the function, first convert the function into standard form of hyperbolas.

$f\left(x\right)=\sqrt{16+4x^2}$

Let us assume$f\left(x\right)=y$

$y=\sqrt{16+4x^2}$

Squaring both sides,

$y^2=16+4x^2$

$y^2-4x^2=16$

Divide by $16$,

localid="1647164055593" $\frac{y^2}{16}-\frac{x^2}{4}=1$

This is the standard form of hyperbola.

Equation of a Hyperbola: Transverse axis along the$y$-axis

An equation of the hyperbola with center at $\left(0,0\right)$, foci at localid="1647163982827" $\left(0,-c\right)$and localid="1647163985722" $\left(0,c\right)$,

and vertices at $\left(0,-a\right)$and localid="1647163978145" $\left(0,a\right)$is

localid="1647163974201" $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$; where$b^2=c^2-a^2$

The transverse axis is the $y$-axis.

Here the equation of the hyperbola is

$\frac{y^2}{16}-\frac{x^2}{4}=1$

It can be written as,

$\frac{y^2}{4^2}-\frac{x^2}{2^2}=1$

Comparing general form,

$a=4,b=2$

Since $c^2=a^2+b^2$,

$c^2=16+4$

$c^2=20$

$c=2\sqrt{5}$

Thus the foci are $\left(0,-c\right)≡\left(0,-2\sqrt{5}\right)$and $\left(0,c\right)≡\left(0,2\sqrt{5}\right)$

The vertices are $\left(0,-a\right)≡\left(0,-4\right)$and $\left(0,a\right)≡\left(0,4\right)$

Here, the transverse axis is$y$-axis.

The graph is