Given:
The exponential function,
[tex]y=ab^x+c[/tex]
It passes through the points (0, 5) and (1, 14) and has a horizontal asymptote at y=2.
To find f(3):
Let us first find the values of a, b and c.
As we know, A function of the form
[tex]f\mleft(x\mright)=a(b^x)+c[/tex]
it always has a horizontal asymptote at y = c
So, substitute the point (0,5) and c = 2.
We get
[tex]\begin{gathered} 5=ab^0+2 \\ 5=a+2 \\ a=3\ldots\ldots\ldots(1) \end{gathered}[/tex]
Substitute the point (1, 14), a=3 and c=2, we get
[tex]\begin{gathered} 14=3(b^1)+2 \\ 3b=12 \\ b=4\ldots\ldots\ldots(2) \end{gathered}[/tex]
So, the given equation becomes,
[tex]f(x)=3(4^x)+2[/tex]
Next, substitute x=3, we get
[tex]\begin{gathered} f(3)=3(4^3_{})+2 \\ =3(64)+2 \\ =192+2 \\ f(3)=194 \end{gathered}[/tex]
Hence, the solution is 194.
