Question

How do I find the holes for this function?

Answer 1

You need to find common factors in the equation. First, we factorise the denominator:

`{x}^(2) - x - 2 \: = > (x + 1)(x - 2)`
Which leaves is with the equation:

`y = ((x - 5)(x + 1)(x - 2))/((x + 1)(x - 2))`
We can see that we have the common factors of (x+1) and (x-2), so these cancel out in the numerator and denominator:

`y = ((x - 5))/(1)`
a.k.a: y = x - 5
So if x+1 and x-2 cancel out, then the x values of the holes are:
x+1=0
x= -1
And
x-2=0
x= 2
Now we plug in each of these numbers into our simplified equation:
y = x - 5
y=(-1)-5
y= -6
This gives us the coordinate (-1,-6)
y = x - 5
y=(2)-5
y= -3
This gives us the coordinate (2,-3)
So the holes are at (-1,-6) and (2,3)

Answer 2

A "hole" is created when a factor can be crossed out of the numerator and denominator.

x^2 - x - 2 factors into (x - 2)(x + 1) both of which can be crossed out or canceled with the factors in the numerator. That creates very really tiny holes at x = 2 and x = - 1.