What is the vertex of #y= x^2+16x-1 #?

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Answer 1 · 2016-02-27T12:49:28

Put the equation into vertex form to find that the vertex is at #(-8, -65)#

The vertex form of a quadratic equation is

#y = a(x-h)^2+k#

and the vertex of that graph is #(h, k)#

To obtain the vertex form, we use a process called completing the square. Doing so in this case is as follows:

#y=x^2+16x-1#

#=x^2+16x+64-65#

#=(x+8)^2-65#

#=(x-(-8))^2-65#

Thus the vertex is at #(-8, -65)#