What is the vertices f(x)=-0.01x^2+0.7x+6 ?

The vertices f(x)=-0.01x^2+0.7x+6 is Maximum (35,18.25)

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vertices f(x)=-0.01x^2+0.7x+6

vertices

f (x) = - 0.01 x^{2} + 0.7 x + 6

Maximum (35, 18.25)

Solution steps

y = - 0.01 x^{2} + 0.7 x + 6

The parabola parameters are:

a = - 0.01, b = 0.7, c = 6

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

x_{v} = - \frac{0.7}{2 ( - 0.01 )}

Simplify

- \frac{0.7}{2 ( - 0.01 )} : 35

x_{v} = 35

Plug in

x_{v} = 35

to find the

y_{v}

value

y_{v} = 18.25

Therefore the parabola vertex is

(35, 18.25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 0.01

Maximum (35, 18.25)

y = + x^{2} - 3

f (x) = (x - 1)^{2} - 4

x^{2} - x - 20

x^{2} - 8 x + 9

x^{2} - 8 x + 7

What is the vertices f(x)=-0.01x^2+0.7x+6 ?
The vertices f(x)=-0.01x^2+0.7x+6 is Maximum (35,18.25)