What is the vertices t^2+2 ?

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

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vertices t^2+2

vertices

t^{2} + 2

Minimum (0, 2)

Solution steps

y = t^{2} + 2

The parabola parameters are:

a = 1, b = 0, c = 2

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

t_{v} = - \frac{0}{2 \cdot 1}

Simplify

t_{v} = 0

Plug in

t_{v} = 0

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(0, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 2)

y = - x^{2} - 4 x + 1

x^{2} - 3 x + 5

3 x^{2} - 3 x

y = 2 x^{2} + 12 x + 17

4 x^{2} + 8 x + 4