What is the vertices t^2 ?

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

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vertices t^2

vertices

t^{2}

Minimum (0, 0)

Solution steps

y = t^{2}

The parabola parameters are:

a = 1, b = 0, c = 0

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} = 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 = 1

Minimum (0, 0)

f (x) = - x^{2} - 6 x + 3

x^{2} - 2 x - 3

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

(x + 5)^{2}

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