What is the vertices x^2-7x-8 ?

The vertices x^2-7x-8 is Minimum (7/2 ,-81/4)

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vertices x^2-7x-8

vertices

x^{2} - 7 x - 8

Minimum (\frac{7}{2}, - \frac{81}{4})

Solution steps

y = x^{2} - 7 x - 8

The parabola parameters are:

a = 1, b = - 7, c = - 8

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

x_{v} = - \frac{( - 7 )}{2 \cdot 1}

Simplify

x_{v} = \frac{7}{2}

Plug in

x_{v} = \frac{7}{2}

to find the

y_{v}

value

y_{v} = - \frac{81}{4}

Therefore the parabola vertex is

(\frac{7}{2}, - \frac{81}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{7}{2}, - \frac{81}{4})

0 = (y - 1)^{2} - 16 (x + 4)

y = + 8 - 9 x^{2}

2 π r^{2}

f (x) = - 3 x^{2} + 6 x

y = 14 (x - 4) (x + 10)