How do you find the vertex and axis of symmetry of $y = x^{2} + 2 x - 1$ $y=x^2+2x-1$ ?

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Answer 1 · 2016-05-06T13:31:37

see below

Axis of symmetric= $- \frac{b}{2 a}$ $-b/(2a)$ = $- \frac{2}{2 (1)}$ $-2/(2(1))$ = $- 1$ $-1$

To find value of y, substitute value of x into equation.

$y = {(- 1)}^{2} + 2 (- 1) - 1$ $y=(-1)^2+2(-1)-1$ = $- 2$ $-2$

Vertex of symmetric is $(- 1, - 2)$ $(-1,-2)$

How do you find the vertex and axis of symmetry of y=x^2+2x-1y=x2+2x−1y=x^2+2x-1?