Graph each function by hand. Be sure to label any intercepts.

$f\left(x\right)=-\sqrt{-25+x^2}$

The function is given by

$f\left(x\right)=-\sqrt{-25+x^2}$

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

$f\left(x\right)=-\sqrt{-25+x^2}$

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

So, $y=-\sqrt{-25+x^2}$

Squaring both sides,

$y^2=-25+x^2$

Substract $x^2$from both sides,

$y^2-x^2=-25$

Multiplying both sides with$\left(-1\right)$

$-y^2+x^2=25$

Rearranging,

$x^2-y^2=25$

Divide by $25$,

$\frac{x^2}{25}-\frac{y^2}{25}=1$

This is standard from of the hyperbolas.

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

An equation of the hyperbola with center at $\left(0,0\right)$, foci at role="math" localid="1647166249992" $\left(-c,0\right)$and role="math" localid="1647166257563" $\left(c,0\right)$, and vertices at role="math" localid="1647166266445" $\left(-a,0\right)$and role="math" localid="1647166275493" $\left(a,0\right)$is

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

The transverse axis is the$x$-axis.

Here, the equation of the hyperbola is

$\frac{x^2}{25}-\frac{y^2}{25}=1$

Also, $\frac{x^2}{5^2}-\frac{y^2}{5^2}=1$

So, $a=5,b=5$

Since role="math" $c^2=a^2+b^2$,

$c^2=25+25$

$c^2=50$

$c=5\sqrt{2}$

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

The vertices are $\left(-a,0\right)\equiv \left(-5,0\right)$and $\left(a,0\right)\equiv \left(5,0\right)$

The transverse axis is the $x$-axis.

The graph is