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`f(x) = (x^2-5)/(x+3)`
(a) Note that a rational function has a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator. Otherwise, the rational function has no horizontal asymptote.
In the above function, the degree of the numerator is 2. This is greater than the degree of the denominator which is 1.
Hence, f(x) has no horizontal asymptote.
(b) A rational function has a vertical asymptote if there are values of x result to zero denominator.
So to solve the vertical asymptote, set the denominator equal to zero.
`x+3 = 0`
`x = -3`
Thus. f(x) has a vertical asymptote which is `x=-3` .
(c) Oblique asymptote exists in a rational function if the degree of the numerator is greater than the degree of the denominator.
As previously motioned in (a), the degree of the numerator of the given function is greater than that of the denominator. So, f(x) has an oblique asymptote.
To solve, expand f(x). Do the long division.
` x` `-` `3`
`x+3` `|bar(x^2 - 0x - 5)`
`-3x - 5`
`(-)` `-3x - 9`
So `f(x) = (x^2-5) / (x+3) = x-3 + 4/(x+3)` .
Note thta the whole part of the division is the oblique asymptote.
Hence, the oblique asymptote of f(x) is `y=x-3` .
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