SOLUTION: If the graph of y=(ax+b)/(x+c) has a horizontal asymptote y=2 and a vertical asymptote x=-3, then a+c=?
a. -5
b. -1
c. 0
d. 1
e. 5
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-> SOLUTION: If the graph of y=(ax+b)/(x+c) has a horizontal asymptote y=2 and a vertical asymptote x=-3, then a+c=?
a. -5
b. -1
c. 0
d. 1
e. 5
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Question 778476: If the graph of y=(ax+b)/(x+c) has a horizontal asymptote y=2 and a vertical asymptote x=-3, then a+c=?
a. -5
b. -1
c. 0
d. 1
e. 5 Found 2 solutions by stanbon, lwsshak3:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If the graph of y=(ax+b)/(x+c) has a horizontal asymptote y=2 and a vertical asymptote x=-3, then a+c=?
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The HA = a/1 = a = 2
----
The VA = -c = -3
----
So a + c = 2+3 = 5
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Cheers,
Stan H.
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You can put this solution on YOUR website! If the graph of y=(ax+b)/(x+c) has a horizontal asymptote y=2 and a vertical asymptote x=-3, then a+c=?
a. -5
b. -1
c. 0
d. 1
e. 5
***
To find the vertical asymptote, set denominator=0, then solve for x:
x+c=0
x=-c=-3 (vertical asymptote)
c=3
..
If degree of numerator=degree of denominator, as in this case:
Horizontal asymptote= quotient of lead coefficient of numerator divided by lead coefficient of denominator.
horizontal asymptote=a/1=2
a=2
a+c=5
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