SOLUTION: please help with these two problems, it's form my college math teacher who's throwing a test this evening please help me. (1) if h(x) =f(x)=2, what would the coordinates of p be

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: please help with these two problems, it's form my college math teacher who's throwing a test this evening please help me. (1) if h(x) =f(x)=2, what would the coordinates of p be      Log On


   

Question 123936: please help with these two problems, it's form my college math teacher who's throwing a test this evening please help me.
(1) if h(x) =f(x)=2, what would the coordinates of p be after the ,give answer in (x,y) form.please show how you did it
(2) if k(x) =f(-x) what would the new coordinates of p be after the reflection?
give answer in (x,y) form please show details
i'm sorry i found one more question,find the equation of the horizontal and vertical asymptoes for the following:
(1) f(x)=2x+3/x+2 and
(2)g(x)=5x/x^2+1, please give horizontal and vertical answers. thanks a lot, i appreciate this.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Sorry, can't help with the first two. You provided insufficient information. You didn't provide the operation that is happening in the first problem, and you didn't give the original coordinates of p in either problem.

As for the other two problems, read on: 





Discussion







Horizontal Asymptotes





There is never more than one horizontal asymptote.





Compare the degree of the numerator polynomial to the degree of the denominator

polynomial.





If the degree of the denominator is larger than the degree of the numerator,

then the horizontal asymptote is always y=0.  In other words, the 

function is asymptotic to the x-axis.





If the degree of the denominator is smaller than the degree of the numerator,

then there is no horizontal asymptote.  There is, however, a slant asymptote.

To determine the slant asymptote, use polynomial long division.  The equation

of the slant asymptote is equal to the quotient excluding the remainder.





If the degree of the denominator equals the degree of the numerator, the

horizontal asymptote is at y = a where a is the quotient of the lead

coefficient on the numerator divided by the lead coefficient on the

denominator.





Vertical asymptotes





There is one vertical asymptote for every real zero of the denominator

polynomial.





Set the denominator polynomial equal to zero and solve for all real roots.





The vertical asymptotes are then x=a%5B1%5D, x=a%5B2%5D, ..., x=a%5Bn%5D





Where a%5B1%5D through a%5Bn%5D are all of the real roots of the 

denominator polynomial.










Solution







I'm going to show a solution for your first problem, then you should be able

to solve the other one.





f%28x%29=%282x%2B3%29%2F%28x%2B2%29





Horizontal:  The degree of the numerator is 1.  The degree of the denominator

is 1.  The lead coefficient on the numerator is 2 and the lead coefficient on

the denominator is 1.  2%2F1=2





Therefore, the horizontal asymptote is y=2





Vertical asymptote:  Set the denominator equal to zero and solve.





x%2B2=0

x=-2





Therefore the vertical asymptote is x=-2.  Since the denominator

polynomial is of degree 1, we know that there is only one real zero of the

polynomial and therefore only one vertical asymptote.










Check Answer







graph%28600%2C600%2C-5%2C5%2C-5%2C5%2C%282x%2B3%29%2F%28x%2B2%29%29





The graphing system on this site doesn't handle functions with vertical

asymptotes very well, so ignore the sort of vertical line on the graph

above.  Note that there is a vertical line at x=-2 that the graph

does not intersect, and a horizontal line at y=2 that is approached

by the function as the absolute value of x gets very large.