What is the vertices f(x)=2(x-5)(x-1) ?

The vertices f(x)=2(x-5)(x-1) is Minimum (3,-8)

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vertices f(x)=2(x-5)(x-1)

vertices

f (x) = 2 (x - 5) (x - 1)

Minimum (3, - 8)

Solution steps

y = 2 (x - 5) (x - 1)

y = 2 (x - 5) (x - 1)

The parabola parameters are:

a = 2, m = 5, n = 1

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

x_{v} = \frac{5 + 1}{2}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(3, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (3, - 8)

- x^{2} + 4 x - 6

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

y^{2} - 6 y - 4 x + 5 = 0

5 x (2 x + 4) - 3 x

f (x) = - 4 x^{2} - 12 x + 5

What is the vertices f(x)=2(x-5)(x-1) ?
The vertices f(x)=2(x-5)(x-1) is Minimum (3,-8)