What is the vertices f(x)=-2x^2+20x+48 ?

The vertices f(x)=-2x^2+20x+48 is Maximum (5,98)

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vertices f(x)=-2x^2+20x+48

vertices

f (x) = - 2 x^{2} + 20 x + 48

Maximum (5, 98)

Solution steps

y = - 2 x^{2} + 20 x + 48

The parabola parameters are:

a = - 2, b = 20, c = 48

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

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

Simplify

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

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 98

Therefore the parabola vertex is

(5, 98)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (5, 98)

y = x^{2} - 15 x

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

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

f (x) = - 0.02 x^{2} + x + 6

y = x^{2} - 1

What is the vertices f(x)=-2x^2+20x+48 ?
The vertices f(x)=-2x^2+20x+48 is Maximum (5,98)