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

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Answer 1 · 2018-02-07T14:59:28

(-6, 33)

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

$y = x^{2} - 4 x + 4 + 16 x - 1$ $y=x^2-4x+4+16x-1$ is the new equation.

Combining like terms, we get

$y = x^{2} + 12 x + 3$ $y=x^2+12x+3$ .

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

$y = {(x + 6)}^{2} - 33$ $y=(x+6)^2-33$ .

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

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

Yay!

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