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
$$f(x) = \frac{rx+s}{2x+t
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-> SOLUTION: A function $f$ has horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$
Part (b): Let $f$ be of the form
$$f(x) = \frac{rx+s}{2x+t
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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) (Show Source):
Igor's input box doesn't understand TeX or LaTeX. Your dollar signs just make your input difficult to read. Use standard plain text notation when posting here, please. See Formatting Math as Text for more information about how to communicate math expressions in plain text that everyone understands.
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