What is the vertices y=12x(x-4) ?

The vertices y=12x(x-4) is Minimum (2,-48)

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vertices y=12x(x-4)

vertices

y = 12 x (x - 4)

Minimum (2, - 48)

Solution steps

y = 12 x (x - 4)

y = 12 x (x - 4)

The parabola parameters are:

a = 12, m = 0, n = 4

x_{v} = \frac{m + n}{2}

x_{v} = \frac{0 + 4}{2}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 48

Therefore the parabola vertex is

(2, - 48)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 12

Minimum (2, - 48)

(x + 2)^{2}

y^{2} + 4 x - 2 y + 21 = 0

f (x) = 415 x^{2} - 4600 x + 200000

y = x^{2} - 4 x 3

f (x) = x^{2} - 3 x - 10