What is the vertices-5x^2+118x+7 ?

The vertices-5x^2+118x+7 is Maximum (59/5 , 3516/5)

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vertices-5x^2+118x+7

vertices

- 5 x^{2} + 118 x + 7

Maximum (\frac{59}{5}, \frac{3516}{5})

Solution steps

y = - 5 x^{2} + 118 x + 7

The parabola parameters are:

a = - 5, b = 118, c = 7

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

x_{v} = - \frac{118}{2 ( - 5 )}

Simplify

- \frac{118}{2 ( - 5 )} : \frac{59}{5}

x_{v} = \frac{59}{5}

Plug in

x_{v} = \frac{59}{5}

to find the

y_{v}

value

y_{v} = \frac{3516}{5}

Therefore the parabola vertex is

(\frac{59}{5}, \frac{3516}{5})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (\frac{59}{5}, \frac{3516}{5})

y^{2} = - 6 x

4 x - 2 x^{2}

y = \frac{95 x ^{2} - 10 x + 26}{10}

(x - 1)^{2} = - 2 y - 11

f (x) = x^{2} - 6 x + 10

What is the vertices-5x^2+118x+7 ?
The vertices-5x^2+118x+7 is Maximum (59/5 , 3516/5)