Question 681992
f(x) = {{{x^2-2x-8}}}
=(x-4)(x+2)
roots/zeros/intercepts at 4 and -2
Plot (4, 0) and (-2, 0)
...
complete the square to find turning point
x^2 - 2x - 8 = 0
x^2 - 2x + 1 = 8 + 1
x^2 - 2x + 1 = 9
y = (x-1)^2 - 9
vertex/turning point at 1, -9
Plot (1,-9)
...
{{{graph(300,200,-10,10,-10,10,x^2-2x-8)}}}
...
f(x) = {{{x^2-2x-15}}}
= (x-5)(x+3)
roots/zeros/intercepts at 5 and -3
Plot (5, 0) and (-3, 0)
...
complete the square to find the turning point
x^2 - 2x - 15 = 0
x^2 - 2x = 15
x^2 - 2x + 1 = 15 + 1
(x-1)^2 - 16 = 0
y = (x-1)^2 - 16
Plot (1,-16)
...
{{{graph(300,200,-10,10,-20,10,x^2-2x-15)}}}
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