What is the vertices f(x)=-(x^2+7x-8) ?

The vertices f(x)=-(x^2+7x-8) is Maximum (-7/2 , 81/4)

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vertices f(x)=-(x^2+7x-8)

vertices

f (x) = - (x^{2} + 7 x - 8)

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

Solution steps

y = - (x^{2} + 7 x - 8)

Rewrite

y = - (x^{2} + 7 x - 8)

in the form

y = a x^{2} + b x + c

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 ( - 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

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

- x^{2} + 2 x - 3

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

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

f (x) = 8 x^{2} + 4 x - 1

f (x) = - x^{2} - 14 x - 48

What is the vertices f(x)=-(x^2+7x-8) ?
The vertices f(x)=-(x^2+7x-8) is Maximum (-7/2 , 81/4)