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

The vertices f(x)=(2x+2)(x-5) is Minimum (2,-18)

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

vertices

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

Minimum (2, - 18)

Solution steps

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

Rewrite

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

in the form

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

y = 2 x^{2} - 8 x - 10

The parabola parameters are:

a = 2, b = - 8, c = - 10

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

x_{v} = - \frac{( - 8 )}{2 \cdot 2}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 18

Therefore the parabola vertex is

(2, - 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (2, - 18)

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

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

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

f (x) = x^{2} - 6 x + 4

x^{2} + x - 4

What is the vertices f(x)=(2x+2)(x-5) ?
The vertices f(x)=(2x+2)(x-5) is Minimum (2,-18)