How do you find the vertex for #2x^-16x+4#?

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Answer 1 · 2015-06-27T21:50:23

I think you mean #2x^2-16x+4#

#2x^2-16x+4 = 2(x-4)^2-28#

This is in vertex form allowing us to read the vertex as #(4, -28)#

Let #f(x) = 2x^2-16x+4#

Then #f(x) = 2(x^2-8x+16)-32+4 = 2(x-4)^2-28#

#f(x) = 2(x-4)^2-28# is in vertex form, being of the form:

#f(x) = a(x-x_1)+y_1#,

where #(x_1, y_1) = (4, -28)# is the vertex and #a# is some constant.

More explicitly:

The minimum value of #f(x)# occurs when #(x-4) = 0#, that is when #x = 4#.

#f(4) = 2(4-4)^2-28 = 0-28 = -28#

So the vertex is at #(4, -28)#