# How do you graph  (y+1)^2/36 - (x+5)^2/9 =1?

##### 1 Answer
May 24, 2017

$\text{center " = (-5, -1); " "a = 6; " } b = 3$
$\text{vertices } \left(- 5 , 5\right) , \left(- 5 , - 7\right)$

#### Explanation:

Given: ${\left(y + 1\right)}^{2} / 36 - {\left(x + 5\right)}^{2} / 9 = 1$

Form is a Vertical Transverse axis hyperbola: ${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1$

$\text{center } = \left(- 5 , - 1\right)$

a = sqrt(36) = 6; " " b = sqrt(9) = 3

Since the larger number in the denominator is over the $y$ value, this is the $a$. Create a dashed box that is $a = 6$ above and below the center in the $y$ direction. Make it be $b = 3$ to the right and left of the center.

Create dashed lines that connect the corners (asymptotes) and extend past the corners. The equations of the asymptotes are: $y = \pm \frac{a}{b} \left(x - h\right) + k$ for the vertical transverse hyperbola.

Let the vertices be at $\left(h , k \pm 6\right) = \left(- 5 , 5\right) , \left(- 5 , - 7\right)$

The hyperbola has two branches. Each branch goes through one vertex and stays inside the asymptotes. 