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

The vertices f(x)=(x-5)(x+1) is Minimum (2,-9)

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

vertices

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

Minimum (2, - 9)

Solution steps

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

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

The parabola parameters are:

a = 1, m = 5, n = - 1

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

x_{v} = \frac{5 + ( - 1 )}{2}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 9)

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

x^{2} - 6 x + 5 y = - 34

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

y + x^{2} = 4 x

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

What is the vertices f(x)=(x-5)(x+1) ?
The vertices f(x)=(x-5)(x+1) is Minimum (2,-9)