What is the vertices (x^3-1)/(x-1) ?

The vertices (x^3-1)/(x-1) is Minimum (-1/2 , 3/4)

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vertices (x^3-1)/(x-1)

vertices

\frac{x ^{3} - 1}{x - 1}

Minimum (- \frac{1}{2}, \frac{3}{4})

Solution steps

y = \frac{x ^{3} - 1}{x - 1}

Rewrite

y = \frac{x ^{3} - 1}{x - 1}

in the form

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

y = x^{2} + x + 1

The parabola parameters are:

a = 1, b = 1, c = 1

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

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

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{3}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{3}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{1}{2}, \frac{3}{4})

+ 10 x^{2} - 15 x + 2

f (x) = 4 x^{2} + 8 x - 8

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

f (x) = 4 x^{2} + 8 x + 3

x = 7 y^{2}

What is the vertices (x^3-1)/(x-1) ?
The vertices (x^3-1)/(x-1) is Minimum (-1/2 , 3/4)