What is the vertices-x^2-x+2 ?

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

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vertices-x^2-x+2

vertices

- x^{2} - x + 2

Maximum (- \frac{1}{2}, \frac{9}{4})

Solution steps

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

The parabola parameters are:

a = - 1, b = - 1, c = 2

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

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

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{9}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- \frac{1}{2}, \frac{9}{4})

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

y = + 16 - (20 x + 15)^{2}

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

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

3 x^{2} - x + 5