Asked by Greg - PV
Given the following rational function:
f(x) = (x^2 + 6x - 8) / (x – 5)
(a) state the domain.
(b) find the vertical and horizontal asymptotes, if any.
(c) find the oblique asymptotes, if any.
f(x) = (x^2 + 6x - 8) / (x – 5)
(a) state the domain.
(b) find the vertical and horizontal asymptotes, if any.
(c) find the oblique asymptotes, if any.
Answers
Answered by
MathMate
Given the rational function:
f(x) = (x^2 + 6x - 8) / (x – 5)
a. Domain
The domain of a rational function is all real numbers <i>minus</i> points where the denominator (x-5) become zero. Here the point to be removed is x-5=0, or x=5.
The answer in interval notation would be:
(-&infin,5)∪(5,∞)
which is essentially all real less x=5.
b. Asymptotes
Vertical asymptotes occur when the denominator becomes zero. There is one such asymptote for f(x).
Hint: this point has been identified in part (a) above.
Horizontal asymptotes are limits of f(x) as x→-∞ or x→∞.
If these limits do not exist, there are no horizontal asymptotes.
Hint: Evaluate Lim x→±∞ and see if the limits exist.
3. oblique asymptotes
Oblique asymptotes exist when the leading term of the numerator divided by the leading term of the denominator yields a linear term (i.e. of the form kx), where k≠0.
Here, the leading term of the numerator is x^2, and that of the denominator is x.
The quotient is therefore x^2/x=x (k=1).
The oblique asymptote is therefore y=x.
Follow link below for the graph:
http://imageshack.us/photo/my-images/221/1307663049.png/
f(x) = (x^2 + 6x - 8) / (x – 5)
a. Domain
The domain of a rational function is all real numbers <i>minus</i> points where the denominator (x-5) become zero. Here the point to be removed is x-5=0, or x=5.
The answer in interval notation would be:
(-&infin,5)∪(5,∞)
which is essentially all real less x=5.
b. Asymptotes
Vertical asymptotes occur when the denominator becomes zero. There is one such asymptote for f(x).
Hint: this point has been identified in part (a) above.
Horizontal asymptotes are limits of f(x) as x→-∞ or x→∞.
If these limits do not exist, there are no horizontal asymptotes.
Hint: Evaluate Lim x→±∞ and see if the limits exist.
3. oblique asymptotes
Oblique asymptotes exist when the leading term of the numerator divided by the leading term of the denominator yields a linear term (i.e. of the form kx), where k≠0.
Here, the leading term of the numerator is x^2, and that of the denominator is x.
The quotient is therefore x^2/x=x (k=1).
The oblique asymptote is therefore y=x.
Follow link below for the graph:
http://imageshack.us/photo/my-images/221/1307663049.png/
Answered by
Anonymous
x^2+x-6/x+3
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