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

The vertices f(x)=2x+x^2 is Minimum (-1,-1)

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vertices f(x)=2x+x^2

vertices

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

Minimum (- 1, - 1)

Solution steps

y = 2 x + x^{2}

Rewrite

y = 2 x + x^{2}

in the form

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

y = x^{2} + 2 x

The parabola parameters are:

a = 1, b = 2, c = 0

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 1)

(x + 2)^{2} = - 6 (y - 1)

y = 2 x^{2} - x

- x^{2} + 14 x - 45

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

y = 3 x^{2} - x - 2