SOLUTION: A function $f$ has horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (b): Let $f$ be of the form $$f(x) = \frac{rx+s}{2x+t

Algebra ->  Functions -> SOLUTION: A function $f$ has horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (b): Let $f$ be of the form $$f(x) = \frac{rx+s}{2x+t      Log On


   



Question 1045870: A function $f$ has horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$
Part (b): Let $f$ be of the form
$$f(x) = \frac{rx+s}{2x+t}.$$
Find an expression for $f(x).$

Answer by solver91311(24713) About Me  (Show Source):
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If there is a horizontal asymptote, then the degree of the numerator and denominator polynomials must be equal (which is true in this case) and the equation of the asymptote must be where is the lead coefficient of the numerator polynomial and is the lead coefficient of the denominator polynomial.

So if your function is:



and if the equation of the horizontal asymptote is



Then



and



A vertical asymptote exists where the denominator polynomial is equal to zero, so if the vertical asymptote is



then



which is to say



In order for the function value to be zero, the numerator must be zero. So, given an x-intercept of



or



Putting it all together,



John

My calculator said it, I believe it, that settles it