What you said about y-intercepts and horizontal asymptotes is correct. Look.
If my exponential function is:
y=5(2/3)^x +1, my graphing calculator shows that its asymptote is y=1 ans its y-intercept is 11.
However, I thought that when you substituted x for 0 (because x=0 at the y-intercept) that you would get the y-intercept from the equation. Well, in doing so, I obtain 6 as my y-intercept.
Anyoe know what's going on?
Thanks in advance to anyone who reads this/answers this/even attempts to answer this.