What is the vertices x^2+8x-5 ?

The vertices x^2+8x-5 is Minimum (-4,-21)

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vertices x^2+8x-5

vertices

x^{2} + 8 x - 5

Minimum (- 4, - 21)

Solution steps

y = x^{2} + 8 x - 5

The parabola parameters are:

a = 1, b = 8, c = - 5

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 21

Therefore the parabola vertex is

(- 4, - 21)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 21)

+ 7 x^{2} + 5

y = 6 x^{2} - 3 x 5

3 x^{2} + 5 x - 12

(x - 1) (x - 3)

y = x^{2} - 8 x + 15