What is the vertices y=24x-3x^2 ?

The vertices y=24x-3x^2 is Maximum (4,48)

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vertices y=24x-3x^2

vertices

y = 24 x - 3 x^{2}

Maximum (4, 48)

Solution steps

y = 24 x - 3 x^{2}

Rewrite

y = 24 x - 3 x^{2}

in the form

y = a x^{2} + b x + c

y = - 3 x^{2} + 24 x

The parabola parameters are:

a = - 3, b = 24, c = 0

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

x_{v} = - \frac{24}{2 ( - 3 )}

Simplify

- \frac{24}{2 ( - 3 )} : 4

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = 48

Therefore the parabola vertex is

(4, 48)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (4, 48)

(x - 5)^{2} = 4 (y - 3)

(x - 4)^{2} = 8 y + 16

- 4 x^{2}

x^{2} - 4 x - 7

f (x) = - 3 x^{2} + 4 x - 1