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#