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January 29, 2015

January 29, 2015

Posted by **Pablo** on Thursday, December 2, 2010 at 8:59pm.

I know it is against the rules in jiskha to do the homework for the students but if I have been looking over my notes and I just don't get this.

example of a rational function from my problems

f(x)=-8x-16/x^2-x-12

Please someone help I have like 40 of these and I really want to learn. A link would help as well.

- Pre-Calculus -
**MathMate**, Thursday, December 2, 2010 at 9:48pmFirst you have to make sure you post algebraic expressions correctly.

Recall that multiplication and division have priority over addition and subtraction, so you expression would be interpreted as:

f(x)=(-8x) -(16/x^2) -x -12

which I do not believe is the intention.

An easy rule to remember is whenever you are transcribing an expression involving division or fractions, add parentheses around the numerator and the denominator if there is more than one term in each. So you rational expression would read:

f(x)=(-8x-16) / (x^2-x-12)

and it would be mathematically correct.

Do you have a textbook? If you don't, and if you are serious about learning calculus, borrow one from the library, or buy a used book for 1/4 of the price at this time of the year (even that is quite expensive, unfortunately).

Do a search on Google about functions, and read up about domain and range, the definition of a functions, the fundamental concepts required later on in Calculus. A link could be:

http://en.wikipedia.org/wiki/Function_%28mathematics%29

Assuming now that you know what a function is, f(x) is the notation, and x is the variable which has to lie within limits of values. These values together make the domain.

A rational function typically (but nt always) contains vertical asymptotes. This occurs when the value of x is such that the denominator becomes zero. So they are easy to spot. In the given function, there are two such points, x=x1, and x=x2. Can you spot them?

There is no horizontal asymptote in the given function f(x) because the degree of the numerator is less than that of the denominator. If they are of the same degree, there would be a horizontal asymptote.

Try to sketch the expressions of the numerator, the denominator, and the function itself on the same graph and you will soon recognize their relationships.

Keep up the good work.

- Pre-Calculus -
**Pablo**, Thursday, December 2, 2010 at 9:50pmThank you so much!!

- Pre-Calculus-correction -
**MathMate**, Friday, December 3, 2010 at 6:58amYou're welcome!

Wait a minute...

I goofed about the horizontal asymptote.

When the degree of the numerator is lower than that of the denominator (in the case of f(x) above), there is a horizontal asymptote at y=0. It is important to take the limit of f(x) as x->±∞ and find out if the function approaches the asymptote from the side of +y or -y.

If the degrees are equal, then the horizontal asymptote is at y=Lim x->∞ of f(x), which is simply the quotient of the leading coefficients.

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