What is the vertices f(x)=-9x^2+9x+1 ?

The vertices f(x)=-9x^2+9x+1 is Maximum (1/2 , 13/4)

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vertices f(x)=-9x^2+9x+1

vertices

Maximum (\frac{1}{2}, \frac{13}{4})

Solution steps

y = - 9 x^{2} + 9 x + 1

The parabola parameters are:

a = - 9, b = 9, c = 1

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

x_{v} = - \frac{9}{2 ( - 9 )}

Simplify

- \frac{9}{2 ( - 9 )} : \frac{1}{2}

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = \frac{13}{4}

Therefore the parabola vertex is

(\frac{1}{2}, \frac{13}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 9

Maximum (\frac{1}{2}, \frac{13}{4})

What is the vertices f(x)=-9x^2+9x+1 ?
The vertices f(x)=-9x^2+9x+1 is Maximum (1/2 , 13/4)