What is the vertices f(x)=x^2+6x ?

The vertices f(x)=x^2+6x is Minimum (-3,-9)

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vertices f(x)=x^2+6x

vertices

f (x) = x^{2} + 6 x

Minimum (- 3, - 9)

Solution steps

y = x^{2} + 6 x

The parabola parameters are:

a = 1, b = 6, c = 0

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

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(- 3, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 9)

3 x^{2} + 2 x + 1

3 x^{2} + 8 y = 0

3 x^{2} + 2 x - 3

2 x^{2} + 4 x + 1

- x^{2} + 2 x + 8