Algebra concepts only:
Find the equation of a rational function with zeroes at $2$ and $3$. No horizontal asymptote and a slant asymptote of $y = x$.
If I had been given only 1 zero, for example x = 2, then I would use the parent graph form to set up my equation as follows: $$ y = x + \frac{a}{x-3} $$ making up a vertical asymptote of 3. Then I would substitute in (2,0) and solve for a. I got a = 2.
However, in this example , two x intercepts are listed as a requirement: My next attempt was to set it up in standard form as follows: $$ \frac{(x - 2)(x - 3)}{p}\qquad \text{ with $p$ representing some factor} $$ Now when I expanded the numerator, I get $x^2 - 5x + 6$ I figured out (by guess check, although there appears to be a pattern that the factor should be (x - 5) to have the quotient be x with some remainder.
My question - is there a systematic way to complete this type of problem?
Problem #2: Find the equation of a rational function with a hole at -3, no horizontal asymptote and a slant asymptote of y = x + 1.
My strategy for this one was to ignore the hole (add later) and first set up a transformed parent equation with the correct slant symptote:
$$ y = x + 1 + \frac{1}{x-4} $$ again I randomly chose a vertical asymptote. Then after expanding to standard form I can just multiply top and bottom by (x + 3). Sound right?
Problem #2: No real zeroes, No vertical asymptote, no horizontal asymptote, and a y intercept of 1/3
I'm not sure this is valid, but how about (x^2 + 1)^2/(x^2 + 3) Wish I knew a more systematic way to do this besides just really fitting things in so that they work - seemingly.