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

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Answer 1 · 2016-02-27T12:49:28

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

The vertex form of a quadratic equation is

$y = a {(x - h)}^{2} + k$ $y = a(x-h)^2+k$

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

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

$y = x^{2} + 16 x - 1$ $y=x^2+16x-1$

$= x^{2} + 16 x + 64 - 65$ $=x^2+16x+64-65$

$= {(x + 8)}^{2} - 65$ $=(x+8)^2-65$

$= {(x - (- 8))}^{2} - 65$ $=(x-(-8))^2-65$

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

What is the vertex of y= x^2+16x-1 y=x2+16x−1y= x^2+16x-1 ?