Scribe Post 9/22/08

Class Notes



We started class today with a warm up: we had to find the common denominator for

[(x+1) / (x^2-5x+6)] - [(3x+11) / (x^2-x-6)]

and simplify

[(X^2-1) / (2x+2)] * [4/(x^2-2x+1)]

Next we went over some problems from the packet and disscussed vertical and horizontal asymptotes. Conclusively it was decided that several rules apply when finding the horizontal asymptote (HA) as x is getting damn big.

1. When the highest power of x is in the denominator the HA: y=0. Example: 1/(x^2+3)


2. When the highest powers of x are equal in the numerator and denominator use the coefficients in front of these values to find the HA. Example: (6x^2)/(3x^2+4) HA: y=6/3 or .5


3. When the highest power of x is in the numerator there is no horizontal asymptote, but there is a slant asymptote. Example: x^3/x^2. as x gets bigger the top and bottom continue to grow, so it just keeps going.

For HAs, however, there are some exceptions. For (x^2-9)/(x+3), a rule 3 case, there is no HA but when x=-3 the value is undefined. The graph of this appears to cross this line where the HA would be but there is actually a "hole," explained by the tern removable discontinuity.*plug this into the y= on your calculator to see the graph* Mr. Marchetti explains this using force fields. VAs are like star trek force fields that cannot be penetrated, while HAs are not. Close to zero they are weaker, but get stronger and eventually impenetrable farther away.

Announcements

Quiz tomorrow, wednesday, and Homework * revisions due thursday

Homework

Study for Quiz: Up to but NOT including rational functions, study tables, story sketches, reciprocal family, and algebra (used for tables) Also, due thursday are more problems from the packet, do some from the sections you need more practice on.

Next Scribe is Julia