What is the vertices-(x-30)^2+40 ?

The vertices-(x-30)^2+40 is Maximum (30,40)

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vertices-(x-30)^2+40

vertices

- (x - 30)^{2} + 40

Maximum (30, 40)

Solution steps

y = - (x - 30)^{2} + 40

Rewrite

y = - (x - 30)^{2} + 40

in the form

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

y = - x^{2} + 60 x - 860

The parabola parameters are:

a = - 1, b = 60, c = - 860

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

x_{v} = - \frac{60}{2 ( - 1 )}

Simplify

- \frac{60}{2 ( - 1 )} : 30

x_{v} = 30

Plug in

x_{v} = 30

to find the

y_{v}

value

y_{v} = 40

Therefore the parabola vertex is

(30, 40)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (30, 40)

f (x) = x^{2} + 3 x - 10

f (x) = x^{2} + 6 x + 11

x^{2} - 6 x + 5 y = 34

f (x) = 5 x^{2} + 1

\frac{5 x ^{2}}{2}