What is the vertices-3x^2+30x-8 ?

The vertices-3x^2+30x-8 is Maximum (5,67)

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vertices-3x^2+30x-8

vertices

- 3 x^{2} + 30 x - 8

Maximum (5, 67)

Solution steps

y = - 3 x^{2} + 30 x - 8

The parabola parameters are:

a = - 3, b = 30, c = - 8

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

x_{v} = - \frac{30}{2 ( - 3 )}

Simplify

- \frac{30}{2 ( - 3 )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 67

Therefore the parabola vertex is

(5, 67)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (5, 67)

f (x) = 4 x^{2} - 8 x + 7

f (x) = 2 x^{2} + 5 x + 2

f (x) = 8 x^{2}

y = 3 x - x^{2}

y = 3 x^{2} + 12 x - 3