SOLUTION: Find an exponential function of the form f(x)= ba^x+c with y intercept 2, horizontal asymptote=-2, that passes through the point P(1,4). My answer is f(x)= 2(1.5^x)-2. Thanks for
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Exponential-and-logarithmic-functions
-> SOLUTION: Find an exponential function of the form f(x)= ba^x+c with y intercept 2, horizontal asymptote=-2, that passes through the point P(1,4). My answer is f(x)= 2(1.5^x)-2. Thanks for
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You can put this solution on YOUR website! Find an exponential function of the form f(x)=Y= ba^x+c with y intercept 2,
THAT FOR X=0...Y=2
Y=2=B*A^0+C=B+C...............................I
horizontal asymptote=-2,
THAT IS AS X TENDS TO + OR - INFINITY Y TENDS TO -2
A^X WILL TEND TO ZERO AS X TENDS TO +INFINITY IF A<1 AND
A^X WILL TEND TO ZERO AS X TENDS TO -INFINITY IF A>1..HENCE WE TAKE THAT A^X WILL TEND TO ZERO AS X TENDS TO + OR - INFINITY TO GET
Y=-2=B*0+C...............................II
C=-2
FROM EQN.I....B-2=2..
B=4
that passes through the point P(1,4). My answer is f(x)= 2(1.5^x)-2. Thanks for the help.
Y=4=4*A^1-2..................III
4A=4+2=6
A=6/4=1.5
ANSWER IS F(X)=Y=4*1.5^X-2...GRAPH IS GIVEN BELOW
