Question 134214: Can you please help me? I just need to ensure I am correct:
Find the range of the quadradic function:
f(x)=(x+4)²+8
Thanks Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! f(x)=(x+4)^2 +8
=x^2+8x+16+8
=x^2+8x+24
a=1, b=8, c=24
a(-b/2a)^2+b(-b/2a)+c=the minimum at y.
-b/2a=-8/2=-4
-4^2+8*-4+24=16-32+24=40-32=8
The range= 8<=y
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Ed
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