# How do you find all the critical points to graph x^2/9 - y^2/16 = 1 including vertices?

Given that

${x}^{2} / 9 - {y}^{2} / 16 = 1$

${x}^{2} / {3}^{2} - {y}^{2} / {4}^{2} = 1$

The above equation shows a horizontal hyperbola with horizontal semi axis $a = 3$ & transverse semi-axis $b = 4$

The center of hyperbola $\left(0 , 0\right)$

The vertices of given hyperbola are $\left(\setminus \pm a , 0\right) \setminus \equiv \left(\setminus \pm 3 , 0\right)$ & $\left(0 , \setminus \pm b\right) \setminus \equiv \left(0 , \setminus \pm 4\right)$

Asymptotes of given hyperbola

$y = \setminus \pm \frac{b}{a} x$

$y = \setminus \pm \frac{4}{3} x$

Specify all the four vertices of hyperbola with center at origin & draw both the asymptotes. Then draw both the horizontal branches of hyperbola