Answer: If the function is: f(x) = (x^2−3x−10)/(x−5) The function has no holes.
If the function is f(x) = x^2−3x−10/x−5 then the hole is at x = 0 (the left side goes up, and the right side goes down, as the numerator is a negative number)
Explanation:
I am not shure of what is the function, as you used no parentheses in it, so i will solve it for two different functions:
We have the function:
f(x) = (x^2−3x−10)/(x−5)
We want to find the "hole", this means that we want to find the value of x where the denominator is equal to zero.
If you look at the function, is easy to think that in x = 5 the denominator will be equal to zero, but we need to prove it.
First, we need to see if the numerator part is also zero when x = 5.
so
5^2 - 3*x - 10 = 25 -15 - 10 = 0
Ok, let's find the other root of the quadratic function:
x^2 - 3*x - 10 = (x - 5)*(x - b) = x^2 - (b + 5)*x + (b*5)
then we have: (b + 5) = 3 and b*5 = -10
then b = -2, this means that we can write the numerator part as:
x^2 - 3*x - 10 = (x - 5)*(x + 2)
then our function is:
f(x) = (x - 5)*(x + 2)/(x - 5) = x + 2
This function has no holes.
Now, if the function actually is:
f(x) = x^2−3x−10/x−5
Then we have an x on the denominator of the third term of the equation, this denominator is 0 only when x = 0