Question 65621: I need help to find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).
Does anyone grasp this?
Found 2 solutions by stanbon, venugopalramana:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I need help to find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).
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If y-int = 2, f(0)=ba^0+c
=b+c=2
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If horizontal asymptote is y=-2, then c=-2
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Then b-2=2
b=4
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Now, y=4a^x-2
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If y=4 when x=1,
4=4a^(1)-2
4=4a-2
4a=6
a=3/2
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EQUATION:
y=4(3/2)^x-2
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Cheers,
Stan H.
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Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! I need help to find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).
Does anyone grasp this?
Y=B*A^X+C
IT GOES THROUGH (1,4)...SO
4=B*A^1+C=AB+C............................1
HORIZINTAL ASYMPTOTE IS Y=-2...THAT IS AS X TENDS TO + OR - INFINITY Y TENDS TO -2.
CASE 1 ...AS X TENDS TO + INFINITY Y TENDS TO -2
-2=B*A^(INFINITY)+C......HENCE |A| MUST BE <1 AND THEN A^(INFINITY)=0
-2=C.............................2
Y INTERCEPT IS 2..THAT IS AT X=0,Y=2
2=B*A^0+C=B+C.............3
2=B-2
B=4
FROM 1...
4=4A-2
A=6/4..............NOT POSSIBLE AS IT IS CONTRARY TO OUR BASIC REQUIREMENT IN CASE 1 THAT |A|<1.SO
CASE 2.....AS X TENDS TO - INFINTY, Y TENDS TO -2.
PROCEEDING AS ABOVE
-2=B*A^(-INFINITY)+C......
-2=C.............................2
Y INTERCEPT IS 2..THAT IS AT X=0,Y=2
2=B*A^0+C=B+C.............3
2=B-2
B=4
FROM 1...
4=4A-2
A=6/4 =1.5.......
HENCE EQN. IS
Y=4(1.5)^X-2
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