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

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 (a): Let $f$ be of the form $$f(x) = \frac{ax+b}{x+c      Log On


   

Question 1045671: A function $f$ has horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$
Part (a): Let $f$ be of the form
$$f(x) = \frac{ax+b}{x+c}.$$
Find an expression for $f(x).$
Part (b): Let $f$ be of the form
$$f(x) = \frac{rx+s}{2x+t}.$$
Find an expression for $f(x).$
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
y=%28ax%2Bb%29%2F%28x%2Bc%29
The vertical asymptote means there is a zero in the denominator.
x%2Bc=0
3%2Bc=0
c=-3
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y=%28ax%2Bb%29%2F%28x-3%29
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Using the x-intercept,
0=%28a%281%29%2Bb%29%2F%281%2B3%29
a%2Bb=0
a=-b
So then using the horizontal asymptote,
y=%28a%2Bb%2Fx%29%2F%281-3%2Fx%29
As x gets very large, the two terms with x in the denominator tend to zero.
-4=%28a%2B0%29%2F%281-0%29
a=-4
So then,
b=-%28-4%29
b=4
Finally,
y=%28-4x%2B4%29%2F%28x-3%29
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Do the second problem the same way.