What is the vertices y=3x^2+18 ?

The vertices y=3x^2+18 is Minimum (0,18)

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vertices y=3x^2+18

vertices

y = 3 x^{2} + 18

Minimum (0, 18)

Solution steps

y = 3 x^{2} + 18

The parabola parameters are:

a = 3, b = 0, c = 18

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

x_{v} = - \frac{0}{2 \cdot 3}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 18

Therefore the parabola vertex is

(0, 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (0, 18)

20 x - 4

f (x) = \frac{x + 1}{x - 5}

- 12 x + 3 y = - 9

\frac{3 ( 6 - t )}{( t + 5 ) ( t - 6 )}

- 3 x^{2} + 3