What is the vertices f(x)=-5x^2+20x+60 ?

The vertices f(x)=-5x^2+20x+60 is Maximum (2,80)

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vertices f(x)=-5x^2+20x+60

vertices

f (x) = - 5 x^{2} + 20 x + 60

Maximum (2, 80)

Solution steps

y = - 5 x^{2} + 20 x + 60

The parabola parameters are:

a = - 5, b = 20, c = 60

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

x_{v} = - \frac{20}{2 ( - 5 )}

Simplify

- \frac{20}{2 ( - 5 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 80

Therefore the parabola vertex is

(2, 80)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (2, 80)

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

f (x) = - x^{2} - 14 x - 42

y = 2 x^{2} + 4 x + 6

y = 2 x^{2} + 4 x - 5

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

What is the vertices f(x)=-5x^2+20x+60 ?
The vertices f(x)=-5x^2+20x+60 is Maximum (2,80)