Question

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

Answer 1

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 `(a+b)(a-b)`. This allows us to factor our expression as

`y=(x+6)(x-6)`

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!

Answer 2

The given curve

`y=x^2-36`

`x^2=y+36`

The above curve shows an upward parabola `X^2=4aY` which has

Vertex: `(x=0, y+36=0)\equiv (0, -36)`

Focus: `(x=0, y+36=1/4)\equiv(0, -143/4)`

Axis of symmetry: `x=0`