Question

Create a table of points (x,y) for the following equations.f(x)= 3x ———— x^2-x-2

Answer 1

`f(x)=(3x)/(x^2-x-2)`

Let's factor out the denominator first so that we can avoid the value of x that makes the function irrational.

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

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.

`y=(3(0))/((0-2)(0+1))=(0)/(-2)=0`

Therefore, at x = 0, y = 0.

Assume x = 1

`y=(3(1))/((1-2)(1+1))=(3)/((-1)(2))=(3)/(-2)=-(3)/(2)=-1.5`

Therefore, at x = 1, y = -1.5

Assume x = -3

`y=(3(-3))/((-3-2)(-3+1))=(-9)/(-5(-2))=-(9)/(10)=-0.9`

At x = -3, y = -0.9.

Assume x = -2

`y=(3(-2))/((-2-2)(-2+1))=(-6)/(-4(-2))=(-6)/(8)=-1.5`

At x = -2, y = -1.5 too.

Assume x = 3

`y=(3(3))/((3-2)(3+1))=(9)/(1(4))=(9)/(4)=2.25`

At x = 3, y = 2.25

Assume x = 4,

`y=(3(4))/((4-2)(4+1))=(12)/(2(5))=(12)/(10)=1.2`

At x = 4, y = 1.2

Assume x = -4

`y=(3(-4))/((-4-2)(-4+1))=(-12)/(-6(-3))=-(12)/(18)=-(2)/(3)`

At x = -4, y = -2/3

Let's summarize the points that we have calculated using table.