Question 681993
Given the roots of -8 and 4; Plot (-8,0) and (4,0)
(x+8) and (x-4) = 0
(x+8)(x-4)=0
{{{x^2-4x + 8x - 32}}}=0
f(x) = {{{x^2 + 4x - 32}}}
...
complete the square to find the turning point
{{{x^2 + 4x = 32}}}
{{{x^2 + 4x + 4 = 32 + 4}}}
(x+2)^2 = 36
y = (x+2)^2 - 36
vertex/turning point of (-2,-36)
...
{{{graph(300,200,-10,10,-40,20,x^2+4x-32)}}}
...
Using roots of -3 and 3; Plot (-3,0) and (3, 0)
(x+3) and (x-3)
(x+3)(x-3)=0
{{{x^2+3x-3x-9}}}=0
f(x) = {{{x^2-9}}}
...
turning point (0,-9)
...
{{{graph(300,200,-10,10,-15,10,x^2-9)}}}
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