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

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

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

vertices

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

Minimum (- 1, - 1)

Solution steps

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)

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

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

y^{2} + 6 y + 9 = x + 8

x^{2} + 4 x - 4 y = 0

x^{2} + y^{2} - 4 x + 8 y - 5 = 0