Question

Determine whether f (x) x^2-2x-3/x^2+3x+2 has any holes. if it does, give coordinates

Answer 1

Yes, this function has one hole

This occurs at
`(-1, -4)`

Explanation:

Definition of hole
A hole on a rational function represents the fact that the function approaches the point, but is not actually defined on that precise x value.

• To identify holes first factor out the numerator and denominator.

• If there is a common factor in the numerator and denominator then there is a hole
• Cancel the common factor from both numerator and denominator
• Set the common factor = 0 and find the corresponding value for x

• Use this x value in the factored function to determine the y-value for the hole

Given function

`f(x) = (x^2-2x\:-3)/(x^2+\:3x\:+\:2)\\`

Factor the numerator:

`x^2 - 2x - 3 = (x+1)(x - 3)`

Factor the denominator:

`x^2+\:3x\:+\:2 = (x + 1)(x + 2)`

Hence

`\begin{aligned}f(x) & = (x^2-2x\:-3)/(x^2+\:3x\:+\:2)\\\\& = ( (x+1)(x - 3))/((x + 1)(x + 2))\\\\ &= (x - 3)/(x+2) \end{aligned}`

The common factor is
`(x + 1)`

Set this = 0 to find the hole:

`x + 1 = 0 = > x = -1`

So at
`x = -1` there is a hole.

Substitute this value of x in the factored function to get the y value for the hole

We have

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

`\begin{aligned}f(2) &= (-1 - 3)/(-1 + 2)\\& = (-4)/(1)\\&= -4\end{aligned}`

So the hole occurs at
`(-1, -4)`

Answer 2

hole at (- 1, - 4 )

Explanation:

If f(x) has terms that cancel , on the numerator/ denominator then a discontinuity corresponding to that factor is removable and there is a hole.

f(x) =
`(x^2-2x-3)/(x^2+3x+2)` ← factor numerator and denominator

=
`((x-3)(x+1))/((x+2)(x+1))` ← cancel (x + 1) on numerator/ denominator

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

this indicates that x + 1 = 0 ( or x = - 1) is a removable discontinuity and the graph has a hole in it.

substitute x = - 1 into the remaining f(x) for y- coordinate of hole

f(- 1) =
`(-1-3)/(-1+2)` =
`(-4)/(1)` = - 4

coordinates of hole are (- 1, - 4 )