What is the vertices 4+2x-x^2 ?

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

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vertices 4+2x-x^2

vertices

4 + 2 x - x^{2}

Maximum (1, 5)

Solution steps

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

Rewrite

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

in the form

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

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

The parabola parameters are:

a = - 1, b = 2, c = 4

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

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

Simplify

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

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 5)

y^{2} = 40 x

y^{2} = 16 x - 8

y = - x^{2} + 3

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

y = 5 x^{2} + 13 x - 6