What is the vertices-3x^2-5 ?

The vertices-3x^2-5 is Maximum (0,-5)

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vertices-3x^2-5

vertices

- 3 x^{2} - 5

Maximum (0, - 5)

Solution steps

y = - 3 x^{2} - 5

The parabola parameters are:

a = - 3, b = 0, c = - 5

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

x_{v} = - \frac{0}{2 ( - 3 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(0, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (0, - 5)

f (x) = x^{2} - 8 x - 9

X^{2} + 2 X - 8

f (x) = x^{2} - 4 x - 4

f (x) = 2 x^{2} + 6 x + 1

x^{2} + 10 x + 11