What is the vertices 4t^2 ?

The vertices 4t^2 is Minimum (0,0)

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vertices 4t^2

vertices

4 t^{2}

Minimum (0, 0)

Solution steps

y = 4 t^{2}

The parabola parameters are:

a = 4, b = 0, c = 0

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

t_{v} = - \frac{0}{2 \cdot 4}

Simplify

t_{v} = 0

Plug in

t_{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 = 4

Minimum (0, 0)

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

(x + 1)^{2} + 1

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

- x^{2} - 2 x + 8

y = 4 x^{2} - 8