What is the vertices y=5x^2 ?

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

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vertices y=5x^2

vertices

y = 5 x^{2}

Minimum (0, 0)

Solution steps

y = 5 x^{2}

The parabola parameters are:

a = 5, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 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 = 5

Minimum (0, 0)

f (x) = - (x + 3)^{2} + 5

4 t^{2}

x^{2} - 6 x - 4 y + 5 = 0

(x + 1)^{2} + 1

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