# How do you find all the critical points to graph 4x^2-9y^2=36 including vertices, foci and asymptotes?

Vertices $\left(\pm 3 , 0\right)$
Focii $\left(\pm \sqrt{13} , 0\right)$
Asymptotes equations: $y = \pm \frac{2}{3} x$

#### Explanation:

The equation $4 {x}^{2} - 9 {y}^{2} = 36$ is a hyperbola

The standard form of the equation of a hyperbola is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

For the given $4 {x}^{2} - 9 {y}^{2} = 36$

We divide both sides of the equation by $36$

$\frac{4 {x}^{2} - 9 {y}^{2}}{36} = \frac{36}{36}$

${x}^{2} / 9 - {y}^{2} / 4 = 1$

it follows that

${\left(x - 0\right)}^{2} / \left({3}^{2}\right) - {\left(y - 0\right)}^{2} / \left({2}^{2}\right) = 1$

and by inspection and by comparing to the form above,

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

we have

$a = 3$ and $b = 2$ and
$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{9 + 4} = \sqrt{13}$
$c = \sqrt{13}$

and center $\left(h , k\right) = \left(0 , 0\right)$

Center $\left(h , k\right) = \left(0 , 0\right)$
Vertices at $\left(h + a , k\right) = \left(+ 3 , 0\right)$
and $\left(h - a , k\right) = \left(- 3 , 0\right)$

Focii at $\left(h + c , k\right) = \left(+ \sqrt{13} , 0\right) = \left(3.6 , 0\right)$
and $\left(h - c , k\right) = \left(- \sqrt{13} , 0\right) = \left(- 3.6 , 0\right)$

The equations of the asymptotes are given by the following formulas

$y = \pm \frac{b}{a} \left(x - h\right) + k$

therefore,

$y = \pm \frac{2}{3} \left(x - 0\right) + 0$

we have

$y = \frac{2}{3} x$ and $y = - \frac{2}{3} x$

Kindly see the graphs of the hyperbola $4 {x}^{2} - 9 {y}^{2} = 36$ colored red and asymptotes $y = \frac{2}{3} x$ and $y = - \frac{2}{3} x$ colored blue. You can also see the colored orange dots (Focii) and colored green dots (Vertices).

God bless... I hope the explanation is useful.