How do you identify the important parts of #y + x^2 = 8x − 12# to graph it?

1 Answer
Shwetank Mauria
Mar 31, 2017

Please see below.

Explanation:

This is a typical quadratic equation, which in vertex form appears as #y=a(x-h)^2+k#, which is a parabola with axis of symmetry as #x=h# and vertex at #(h,k)#, where we have a maxima or minima depending on whether #a# is negative or positive.

In such cases, we need to identify

  1. first the axis of symmetry

  2. then the vertex and finally

  3. a few points on both sides of axis of symmetry.

Hence, let us first convert it into vertex form. As we have #y+x^2=8x-12#, let us convert it into vertex form. The equation becomes

#y=-x^2+8x-12#

i.e. #y=-1(x^2-2xx4xx x+4^2-4^2)-12#

or #y=-1(x-4)^2+16-12#

or #y=-1(x-4)^2+4#

Now as #a=-1#, we have a maxima at #(4,4)#

axis of symmetry is #x=4#

Let us choose points with #x={-4,-2,0,2,6,8,10,12}#
(around #x=4#)

Corresponding value of #y# will be #y={-60,-32,-12,0,0,-12,-32,-60}#

and graph appears as follows:
graph{y+x^2=8x-12 [-6, 14, -70, 10]}

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