What is the vertices f(x)=x^2+8x ?

The vertices f(x)=x^2+8x is Minimum (-4,-16)

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vertices f(x)=x^2+8x

vertices

f (x) = x^{2} + 8 x

Minimum (- 4, - 16)

Solution steps

y = x^{2} + 8 x

The parabola parameters are:

a = 1, b = 8, c = 0

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(- 4, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 16)

f (x) = x^{2} + 7

f (x) = x^{2} + 9

y = x^{2} + 12 x + 36

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

f (x) = x (115 - 4 x)