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# HELPS PLEASE! also ty CPhill for helping me on the first one.

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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).

Apr 14, 2019

#1
+100571
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a)  If the vertical asymptote  = 3

Then   this value of x makes the denominator = 0

So......3 + c  = 0 ⇒  c = -3

If we have a horizontal asymptote of - 4...this implies that (ax/1x) = (a / 1) =  - 4  ⇒  a = -4

And if the x intercept =   then   a(1) + b = 0  ...so...

-4(1) + b = 0

-4 + b = 0

b = 4

So   (a, b , c ) =  ( -4, 4, -3)

Apr 14, 2019
#2
+100571
+2

(b)   Similarly

2(3) + t = 0

6 + t = 0

t = - 6

And

(r) / (2) = -4

r = 8

And if we have (1, 0)   then

r(1) + s = 0

8(1) + s = 0

s = -8

So    (r, s, t )  =  ( 8, -8, -6)

Apr 14, 2019
#3
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THANK YOU!!!!!!

Apr 15, 2019
#4
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Also isn't it r=-8 in part b?

Apr 16, 2019