en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=3(x-1)(x+2)

Solution

vertices

f (x) = 3 (x - 1) (x + 2)

Solution

Minimum (- \frac{1}{2}, - \frac{27}{4})

Solution steps

y = 3 (x - 1) (x + 2)

y = 3 (x - 1) (x + 2)

The parabola parameters are:

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

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

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

Simplify

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = - \frac{27}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{27}{4})

If

a < 0,

then the vertex is a maximum value

If

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{1}{2}, - \frac{27}{4})

Popular Examples

vertices f(x)=-1/250 x^2+4/5 x vertices

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x

vertices y=2x^2-8x+4 vertices

y = 2 x^{2} - 8 x + 4

vertices f(x)=4(x+4)^2+6 vertices

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

vertices 4(x+2)^2+5 vertices

4 (x + 2)^{2} + 5

vertices x^2-2x-80 vertices

x^{2} - 2 x - 80

Frequently Asked Questions (FAQ)

What is the vertices f(x)=3(x-1)(x+2) ?
The vertices f(x)=3(x-1)(x+2) is Minimum (-1/2 ,-27/4)