How do you solve the equation #x^2+8=-6x# by graphing?

1 Answer
Oct 2, 2017

See below.

Explanation:

To graph #x^2+8=-6x#

First add #-6x# to both sides to get:

#x^2+6x+8=0#

We can now write this in function form by replacing the #0# with a #y#:

#x^2 +6x+8= y# or. #y=x^2 +6x+8#

Find any intercepts ( not the x axis intercepts, since we are going to find them from the graph):

#y# axis intercepts occur where #x = 0#, so:

#0^2+6(0) + 8 = 8#

So #y# axis intercept at #(0 , 8 )#

We can now find the vertex. This will give us the coordinates of the highest or lowest point on the parabola.

To find the vertex we need to get #y=x^2 +6x+8# into the form #y = a(x-color(blue)(h))^2 + color(red)(k)#.

Where #a# is the coefficient of #x^2# #color(blue)(h)# is the axis of symmetry and #color(red)(k)# is the minimum/maximum value.

To find the vertex:
Bracket off the terms containing the variable:

#(x^2 + 6x ) + 8#

Factor out the coefficient of #x^2# if this is not #1#
Add the square of half the coefficient of #x# inside the bracket and subtract it outside the bracket:

#(x^2 +6x +(3)^2) - (3)^2+8#

Change #(x^2 +6x +(3)^2)# into the square of a binomial:

#(x^2 + 3)^2-(3)^2+8#

Collect terms outside the bracket:

#(x^2 + 3)^2-1#

This shows that #color(blue)(h)= -3# and #color(blue)(k)=-1#

So the coordinate of the vertex is #( -3 , -1 )#

To find other points we will have to put in some values for #x# and calculate the corresponding values of #y#

Examples:

#x = 1=>y= 15#
#x = 2=>y= 24#
#x = -1=>y= 3#
#x = -5=>y = 4#
#x=-7=>y= 15#

All our coordinates including the vertex and #y# intercept are:

#( -3 , -1 )# # ( 1 , 15 )# #( 4 , 24 )# # ( -1 , 3 )# #( -5 , 3 )# #(-7 , 15 )# #( 0 , 8 )#

Plot these points and draw the parabola:
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Join points:

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Roots will be where parabola crosses the #x# axis. This looks to be at #(-4 , 0 )# and #( -2 , 0 )#