A function $f$ has a 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 $f(x).$
+128472 ax + b
_____
x + c
If the vertical asymptote is at x = 3 then c = -3
If the horizontal asymptote is at y = - 4, then a = -4
If (1,0) is an x intercept then we can solve this for b
-4(1) + b = 0
b = 4
So.....the function is
y = -4x + 4
______
x - 3
Here's the graph : https://www.desmos.com/calculator/t5wbj4l2ad
+128472 Second one :
y = rx + s
_____
2x + t
Vertical asymptote ⇒ x = 3
2(3) + t = 0
6 + t = 0
t = -6
Horizontal asymptote ⇒ y = -4
So r/2 = - 4 ⇒ r = -8
And if (1,0) is the x intercept....we have that
-8(1) + s = 0
s = 8
And the function is :
y = -8x + 8
________
2x - 6
Here's the graph : https://www.desmos.com/calculator/bqxafolfmx
