What is the vertices f(x)=x(115-4x) ?

The vertices f(x)=x(115-4x) is Maximum (115/8 , 13225/16)

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vertices f(x)=x(115-4x)

vertices

f (x) = x (115 - 4 x)

Maximum (\frac{115}{8}, \frac{13225}{16})

Solution steps

y = x (115 - 4 x)

y = x (115 - 4 x)

The parabola parameters are:

a = - 4, m = 0, n = \frac{115}{4}

x_{v} = \frac{m + n}{2}

x_{v} = \frac{0 + ( \frac{115}{4} )}{2}

Simplify

\frac{0 + ( \frac{115}{4} )}{2} : \frac{115}{8}

x_{v} = \frac{115}{8}

Plug in

x_{v} = \frac{115}{8}

to find the

y_{v}

value

y_{v} = \frac{13225}{16}

Therefore the parabola vertex is

(\frac{115}{8}, \frac{13225}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4

Maximum (\frac{115}{8}, \frac{13225}{16})

x^{2} + 5 x + 1

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

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

y = - x^{2} + 10 x

f (x) = 12 x - 2 x^{2}

What is the vertices f(x)=x(115-4x) ?
The vertices f(x)=x(115-4x) is Maximum (115/8 , 13225/16)