How do you find the vertex of the parabola: $y=x^2+4x-5$ ?

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Answer 1 · 2015-09-02T15:23:08

Find the vertex of y = x^2 + 4x - 5.

Ans: (-2, -9)

x-coordinate of the vertex:
$x = (-b/2a) = -4/2 = -2$
y-coodinate of vertex: $y = f(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9$

How do you find the vertex of the parabola: y=x^2+4x-5y=x^2+4x-5?