What is the vertices 4t^2+1 ?

The vertices 4t^2+1 is Minimum (0,1)

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vertices 4t^2+1

vertices

4 t^{2} + 1

Minimum (0, 1)

Solution steps

y = 4 t^{2} + 1

The parabola parameters are:

a = 4, b = 0, c = 1

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} = 1

Therefore the parabola vertex is

(0, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (0, 1)

f (x) = \frac{1}{2} (x - 1)^{2}

x^{2} + 1

2 x^{2} - 5 x + 3

2 X^{2} + 3 X - 20

2 X^{2}