Need help finding the range of a function

**krzyrice** · February 16th 2010, 03:25 PM

The problem is h(x) = x^2 + 8x
I know the domain is all real. But did I solve for x correctly?

I have set up the problem:
y = x^2 + 8x
y = x(x + 8)
y/x = x + 8
y/x - 8 = x
y-8 = x^2
x = square root y - 8

But I don't get how they got the answer of x is greater than or equal to -16. Please help me.

**skeeter** · February 16th 2010, 04:52 PM

complete the square ...

$y = x^2 + 8x$

$y + 16 = x^2 + 8x + 16$

$y + 16 = (x + 4)^2$

$y = (x+4)^2 - 16$

the parabola's vertex is located at $(-4, -16)$ , therefore the range is $y \ge -16$

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