# How do I graph (y+3)^2/25-(x+2)^2/16=1 algebraically?

Sep 10, 2014

This is the equation of a hyperbola.

To graph it, you need to interpret your equation to see what goes where.

So the standard hyperbola equations are:

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

Equation 1 represents a horizontal hyperbola (which looks like this): Equation 2 Represents a vertical hyperbola (shown below): So first up, we will need to decide which one we have in the given example.

The given equation is:

${\left(y + 3\right)}^{2} / 25 - {\left(x + 2\right)}^{2} / 16 = 1$

Since the y is the first term, we know that this hyperbola is a vertical hyperbola.

Now there are two things we need to find in order to graph:

1. The center of the hyperbola.
2. How far up & down the branches are.

To find the center, all we need to know is that in any hyperbola equation, the $h$ and the $k$ represent how far left/right and up/down the hyperbola has translated.

Therefore, the coordinate $\left(h , k\right)$ must be the new translated center of the hyperbola.

So in our equation, $h = - 2$

• It's not just 2 because remember, it's $\left(x - h\right)$

Also, $k = - 3$

• Once again, be sure that you've got your signs right.

Therefore, the center is $\left(h , k\right)$ which is (-2, -3)

Now for how far up/down, we simply need to understand the following:

1. $a$ represents how far left and right the vertices of the branches are located.
2. $b$ represents how far up and down the vertices of the branches are located.

The vertices are simply the points where the branches of the hyperbola begin to curve (refer to image below): Because our hyperbola is vertical, we only need to look up and down , and hence need only the b value.

BE CAREFUL HERE:

What is located in the equation is ${b}^{2}$, not $b$.
Therefore we will need to square root it before we can use it.

In this equation, ${b}^{2} = 25$

$b = \sqrt{25} = \pm 5$

Now we add these onto the $k$ value (since it's the "up/down factor" of our center), and we have our vertices

$- 3 \pm 5 = 2 , - 8$