What is the vertices x^2-5y=0 ?

The vertices x^2-5y=0 is Minimum (0,0)

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vertices x^2-5y=0

vertices

x^{2} - 5 y = 0

Minimum (0, 0)

Solution steps

x^{2} - 5 y = 0

Rewrite

x^{2} - 5 y = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{5}

The parabola parameters are:

a = \frac{1}{5}, b = 0, c = 0

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

x_{v} = - \frac{0}{2 ( \frac{1}{5} )}

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 = \frac{1}{5}

Minimum (0, 0)

y = 2 x^{2} - 8 x

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

y = 2 x^{2} - 6 x

f (x) = - \frac{1}{3} x^{2} - \frac{7}{6} x + 1

4 x^{2} + 4 x - 8