What is the vertices y=x^2-4x3 ?

The vertices y=x^2-4x3 is Minimum (6,-36)

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vertices y=x^2-4x3

vertices

y = x^{2} - 4 x 3

Minimum (6, - 36)

Solution steps

y = x^{2} - 4 x \cdot 3

Rewrite

y = x^{2} - 4 x 3

in the form

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

y = x^{2} - 12 x

The parabola parameters are:

a = 1, b = - 12, c = 0

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

x_{v} = - \frac{( - 12 )}{2 \cdot 1}

Simplify

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = - 36

Therefore the parabola vertex is

(6, - 36)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, - 36)

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

- x^{2} + 2 x - 10

f (x) = 3 x^{2} - 12 x + 13

y = x^{2} - x - 6

f (x) = - x^{2} + 6 x - 8