Consider the function f(x) = the quantity x squared minus 25 end quantity over x plus 4. Determine all asymptotes of this function including horizontal, vertical, and oblique (slant).
@KinzaN
Do you know when you have an oblique asymptote in rational functions?
^^^
When the highest degree of the numerator is one more than the highest degree of the denominator. In simpler terms, it's when the biggest exponent on top is one more than the biggest exponent on the bottom, like in this case.
Since you cannot cancel any factors after factoring the numerator, which is a difference of squares, there are no point discontinuities--no holes!
So would be the answer? how would I anser all of the requirements?
Well, there is a vertical asymptote. The denominator cannot be 0, so the x value that would make the denominator equal to 0 cannot exist.
There is a vertical asymptote when x = -4.
To find the oblique asymptote algebraically requires a lot of explanation. But the short-cut is to perform the division. Your quotient will be the equation of the oblique asymptote.
So that would be (x^2 - 25)/(x+4)
Both synthetic and long division will solve the problem.
So what would it be?
and what is the horizontal one?
You cannot have both a horizontal and an oblique asymptote.
It's either one or the other.
But you CAN have both a vertical asymptote and a point discontinuity. It's confusing.
So whats my oblique then?
so far my answer would be There is a vertical asymptote when x = -4. There is no horizontal asymptote and the oblique is .....
@KinzaN hello?
Can you do the division? You must have learned it, lol
Divide the numerator by the denom. and your quotient is the equation of the oblique asymptote.
I get x-4-9/x+4. @KinzaN is that correct?
I'll check for you. :)
Ok thank you
Ah, you don't add the remainder. The quotient is simply x - 4, so the equation of the oblique slant is y = x - 4
oh.. ok thanks
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