What is the vertices 1-x-x^2 ?

The vertices 1-x-x^2 is Maximum (-1/2 , 5/4)

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vertices 1-x-x^2

vertices

1 - x - x^{2}

Maximum (- \frac{1}{2}, \frac{5}{4})

Solution steps

y = 1 - x - x^{2}

Rewrite

y = 1 - x - x^{2}

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

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

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{5}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{5}{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{5}{4})

f (x) = x^{2} - 5 x + 4

x^{2} + 4

y = x^{2} - 4 x - 5

y^{2} - 2 y - 8 x - 31 = 0

x^{2} - x - 6