What is the vertices f(x)=(x+2)(x+3) ?

The vertices f(x)=(x+2)(x+3) is Minimum (-5/2 ,-1/4)

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vertices f(x)=(x+2)(x+3)

vertices

f (x) = (x + 2) (x + 3)

Minimum (- \frac{5}{2}, - \frac{1}{4})

Solution steps

y = (x + 2) (x + 3)

y = (x + 2) (x + 3)

The parabola parameters are:

a = 1, m = - 2, n = - 3

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

x_{v} = \frac{( - 2 ) + ( - 3 )}{2}

Simplify

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{5}{2}, - \frac{1}{4})

y^{2} = 6 x - 4 y - 12

y = + x^{2} + 8

- 3 x^{2} + 30 x - 8

f (x) = 4 x^{2} - 8 x + 7

f (x) = 2 x^{2} + 5 x + 2

What is the vertices f(x)=(x+2)(x+3) ?
The vertices f(x)=(x+2)(x+3) is Minimum (-5/2 ,-1/4)