What is the vertices x^2-8x ?

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

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vertices x^2-8x

vertices

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)

\frac{40 x ^{3}}{3 x}

f (x) = 3 x^{2} + 6 x + 1

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

f (x) = 3 x^{2} + 6 x - 1

t^{2} + 2