What is the vertices y=x^2+4x-5 ?

The vertices y=x^2+4x-5 is Minimum (-2,-9)

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vertices y=x^2+4x-5

vertices

y = x^{2} + 4 x - 5

Minimum (- 2, - 9)

Solution steps

y = x^{2} + 4 x - 5

The parabola parameters are:

a = 1, b = 4, c = - 5

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(- 2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 9)

y = \frac{- 3}{12 - x - x ^{2}}

y = \frac{x}{x - 2}

f (x) = \frac{7}{4 - 2 x}

g (x) = \frac{2 x}{x ^{2} + 2 x - 24}

f (x) = 2 x^{2} + 8 x - 34