What is the vertices f(x)=-1/250 x^2+4/5 x ?

The vertices f(x)=-1/250 x^2+4/5 x is Maximum (100,40)

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vertices f(x)=-1/250 x^2+4/5 x

vertices

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x

Maximum (100, 40)

Solution steps

y = - \frac{1}{250} x^{2} + \frac{4}{5} x

Rewrite

y = - \frac{1}{250} x^{2} + \frac{4}{5} x

in the form

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

y = - \frac{x ^{2}}{250} + \frac{4 x}{5}

The parabola parameters are:

a = - \frac{1}{250}, b = \frac{4}{5}, c = 0

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

x_{v} = - \frac{( \frac{4}{5} )}{2 ( - \frac{1}{250} )}

Simplify

- \frac{\frac{4}{5}}{2 ( - \frac{1}{250} )} : 100

x_{v} = 100

Plug in

x_{v} = 100

to find the

y_{v}

value

y_{v} = 40

Therefore the parabola vertex is

(100, 40)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{250}

Maximum (100, 40)

y = 2 x^{2} - 8 x + 4

f (x) = 4 (x + 4)^{2} + 6

4 (x + 2)^{2} + 5

x^{2} - 2 x - 80

y^{2} + 2 y - 4 x + 9 = 0

What is the vertices f(x)=-1/250 x^2+4/5 x ?
The vertices f(x)=-1/250 x^2+4/5 x is Maximum (100,40)