What is the vertices f(x)=-x^2+7x+18 ?

The vertices f(x)=-x^2+7x+18 is Maximum (7/2 , 121/4)

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vertices f(x)=-x^2+7x+18

vertices

f (x) = - x^{2} + 7 x + 18

Maximum (\frac{7}{2}, \frac{121}{4})

Solution steps

y = - x^{2} + 7 x + 18

The parabola parameters are:

a = - 1, b = 7, c = 18

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

x_{v} = - \frac{7}{2 ( - 1 )}

Simplify

x_{v} = \frac{7}{2}

Plug in

x_{v} = \frac{7}{2}

to find the

y_{v}

value

y_{v} = \frac{121}{4}

Therefore the parabola vertex is

(\frac{7}{2}, \frac{121}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{7}{2}, \frac{121}{4})

y = (2 x - 1) (5 x - 6)

f (x) = x^{2} + 3 x + 4

f (x) = x^{2} + 3 x + 2

x^{2} - 6 x

x^{2} - 8 x

What is the vertices f(x)=-x^2+7x+18 ?
The vertices f(x)=-x^2+7x+18 is Maximum (7/2 , 121/4)