What is the vertices y=(x^2x^1)/x ?

The vertices y=(x^2x^1)/x is Minimum (0,0)

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vertices y=(x^2x^1)/x

vertices

y = \frac{x ^{2} x ^{1}}{x}

Minimum (0, 0)

Solution steps

y = \frac{x ^{2} x ^{1}}{x}

Rewrite

y = \frac{x ^{2} x ^{1}}{x}

in the form

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

y = x^{2}

The parabola parameters are:

a = 1, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 0)

X^{2}

3 - x^{2}

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

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

f (x) = 3 x^{2}