how do I find the minimum of y = x^2 - 4x - 32?

I think it starts with y = a(x)^2 + bx + c a > 0

then, y = - b/2a

how do I find the minimum of y = x^2 - 4x - 32?

I think it starts with y = a(x)^2 + bx + c a > 0

then, y = - b/2a

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Middletown, CT

Hi again L;

y = x^{2} - 4x - 32

As you already know, this is a parabola. The x^{2} is positive. Therefore, it is opening upward. The minimum corresponds to the line-of-symmetry.

This is NOT...

y=-b/2a

This is...

x=-b/2a

x=[(-(-4)]/[(2)(1)]

x=4/2

x=2

y = x^{2} - 4x - 32

y=(2)^{2}-(4)(2)-32

y=4-8-32

y=4-40

y=-36

The lowest point is (2,-36).

An alternate approach is to consider the fact that the minimum corresponds to the point at which the change of y is zero. Therefore, we can take the first derivative of the equation and calculate this...

y = x^{2} - 4x - 32

0=2x-4

4=2x

2=x

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