Let's factor out the denominator first so that we can avoid the value of x that makes the function irrational.
[tex]f(x)=\frac{3x}{(x-2)(x+1)}[/tex]Therefore, we cannot have a value of x = 2 and x = -1, or else the denominator becomes zero.
Now, let's start creating a table of values for this equation.
Assume x = 0, substitute this value of x to the equation to get the value of f(x) or y.
[tex]y=\frac{3(0)}{(0-2)(0+1)}=\frac{0}{-2}=0[/tex]Therefore, at x = 0, y = 0.
Assume x = 1
[tex]y=\frac{3(1)}{(1-2)(1+1)}=\frac{3}{(-1)(2)}=\frac{3}{-2}=-\frac{3}{2}=-1.5[/tex]Therefore, at x = 1, y = -1.5
Assume x = -3
[tex]y=\frac{3(-3)}{(-3-2)(-3+1)}=\frac{-9}{-5(-2)}=-\frac{9}{10}=-0.9[/tex]At x = -3, y = -0.9.
Assume x = -2
[tex]y=\frac{3(-2)}{(-2-2)(-2+1)}=\frac{-6}{-4(-2)}=\frac{-6}{8}=-1.5[/tex]At x = -2, y = -1.5 too.
Assume x = 3
[tex]y=\frac{3(3)}{(3-2)(3+1)}=\frac{9}{1(4)}=\frac{9}{4}=2.25[/tex]At x = 3, y = 2.25
Assume x = 4,
[tex]y=\frac{3(4)}{(4-2)(4+1)}=\frac{12}{2(5)}=\frac{12}{10}=1.2[/tex]At x = 4, y = 1.2
Assume x = -4
[tex]y=\frac{3(-4)}{(-4-2)(-4+1)}=\frac{-12}{-6(-3)}=-\frac{12}{18}=-\frac{2}{3}[/tex]At x = -4, y = -2/3
Let's summarize the points that we have calculated using table.
