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

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Answer 1 · 2018-02-07T14:59:28

(-6, 33)

The graph #y=(x-2)^2+16x-1# can be expanded.

#y=x^2-4x+4+16x-1# is the new equation.

Combining like terms, we get

# y=x^2+12x+3 #.

We can change this into #y=a(x-h)+k# form.

#y=(x+6)^2-33#.

The vertex must be #(-6, -33)#.

To check, here is our graph: graph{y=x^2+12x+3 [-37.2, 66.8, -34.4, 17.64]}

Yay!