What is the vertices y=8x^2+22x-13 ?

The vertices y=8x^2+22x-13 is Minimum (-11/8 ,-225/8)

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vertices y=8x^2+22x-13

vertices

y = 8 x^{2} + 22 x - 13

Minimum (- \frac{11}{8}, - \frac{225}{8})

Solution steps

y = 8 x^{2} + 22 x - 13

The parabola parameters are:

a = 8, b = 22, c = - 13

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

x_{v} = - \frac{22}{2 \cdot 8}

Simplify

- \frac{22}{2 \cdot 8} : - \frac{11}{8}

x_{v} = - \frac{11}{8}

Plug in

x_{v} = - \frac{11}{8}

to find the

y_{v}

value

y_{v} = - \frac{225}{8}

Therefore the parabola vertex is

(- \frac{11}{8}, - \frac{225}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 8

Minimum (- \frac{11}{8}, - \frac{225}{8})

What is the vertices y=8x^2+22x-13 ?
The vertices y=8x^2+22x-13 is Minimum (-11/8 ,-225/8)