Question 44727
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
{{{ graph( 600, 600, -50, 50, -10, 10, 4*(1.5)^x-2 )}}}