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# help

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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 $f(x).$

Mar 9, 2018

#1
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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   Mar 9, 2018
#2
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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   Mar 9, 2018