What is the vertices f(x)=-0.02x^2+x+6 ?

The vertices f(x)=-0.02x^2+x+6 is Maximum (25,18.5)

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vertices f(x)=-0.02x^2+x+6

vertices

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

Maximum (25, 18.5)

Solution steps

y = - 0.02 x^{2} + x + 6

The parabola parameters are:

a = - 0.02, b = 1, c = 6

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

x_{v} = - \frac{1}{2 ( - 0.02 )}

Simplify

- \frac{1}{2 ( - 0.02 )} : 25

x_{v} = 25

Plug in

x_{v} = 25

to find the

y_{v}

value

y_{v} = 18.5

Therefore the parabola vertex is

(25, 18.5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 0.02

Maximum (25, 18.5)

y = x^{2} - 1

+ \frac{- 1}{2} x^{2} + 18

x^{2} - 4 x + 12

y = x^{2} - x

f (x) = 6 x^{2} - 1

What is the vertices f(x)=-0.02x^2+x+6 ?
The vertices f(x)=-0.02x^2+x+6 is Maximum (25,18.5)