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

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Answer 1 · 2018-07-04T20:00:54

See below:

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 · 2018-07-04T20:21:28

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: $x=0$

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