Question 82637
the y-intercept is 2 ... so when x=0, f(x)=ba^0+c=b(1)+c=b+c=2


horiz asymptote is y=-2 ... as x approaches negative infinity, ba^x approaches 0 ... so c=-2 ... b=2-c ... so b=4


substituting point P(1,4) ... 4=4a^1-2 ... 6=4a ... 3/2=a


f(x)=4(3/2)^x-2