What is the vertices t^2+12t-18 ?

The vertices t^2+12t-18 is Minimum (-6,-54)

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vertices t^2+12t-18

vertices

t^{2} + 12 t - 18

Minimum (- 6, - 54)

Solution steps

y = t^{2} + 12 t - 18

The parabola parameters are:

a = 1, b = 12, c = - 18

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

t_{v} = - \frac{12}{2 \cdot 1}

Simplify

t_{v} = - 6

Plug in

t_{v} = - 6

to find the

y_{v}

value

y_{v} = - 54

Therefore the parabola vertex is

(- 6, - 54)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 6, - 54)

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

10 y^{2} = 22 x

2 x^{2} + 5 x - 7

x^{2} + 12 y = 0

- 4.9 t^{2} + 9.8 t