# How do you identify the important parts of y = x^2 − 36 to graph it?

Jul 4, 2018

See below:

#### Explanation:

We know we will be dealing with an upward opening parabola, since the coefficient on the ${x}^{2}$ term is positive.

One thing we can do is factor this expression so we can find its zeroes, or $x$-intercepts.

You might immediately recognize that we're dealing with a difference of squares of the form

${a}^{2} - {b}^{2}$, which factors as $\left(a + b\right) \left(a - b\right)$. This allows us to factor our expression as

$y = \left(x + 6\right) \left(x - 6\right)$

Setting both factors equal to zero, we get

$x = - 6$ and $x = 6$. These are points we can plot, but it might help to find our $y$-intercept. Let's set $x$ equal to zero to get

$y = - 36$, which is our $y$-intercept. Now, we can graph:

graph{x^2-36 [-80, 80, -40, 40]}

Hope this helps!

The given curve

$y = {x}^{2} - 36$

${x}^{2} = y + 36$

The above curve shows an upward parabola ${X}^{2} = 4 a Y$ which has

Vertex: $\left(x = 0 , y + 36 = 0\right) \setminus \equiv \left(0 , - 36\right)$

Focus: $\left(x = 0 , y + 36 = \frac{1}{4}\right) \setminus \equiv \left(0 , - \frac{143}{4}\right)$

Axis of symmetry: $x = 0$