What is the vertices h(t)=4t^2 ?

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

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vertices h(t)=4t^2

vertices

h (t) = 4 t^{2}

Minimum (0, 0)

Solution steps

y = 4 t^{2}

The parabola parameters are:

a = 4, m = 0, n = 0

x_{v} = \frac{m + n}{2}

t_{v} = \frac{0 + 0}{2}

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)

f (x) = 19 x^{4} - 114 x^{2}

f (x) = \frac{3 + 4 x}{3 x}

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

f (x) = \frac{5}{x + 9}

3 x + y = 9