Question

I need to know what was the horizontal asymptote at y=2 and what is f(3)

Answer 1

Given:

The exponential function,

`y=ab^x+c`

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

`f\mleft(x\mright)=a(b^x)+c`

it always has a horizontal asymptote at y = c

So, substitute the point (0,5) and c = 2.

We get

`\begin{gathered} 5=ab^0+2 \\ 5=a+2 \\ a=3\ldots\ldots\ldots(1) \end{gathered}`

Substitute the point (1, 14), a=3 and c=2, we get

`\begin{gathered} 14=3(b^1)+2 \\ 3b=12 \\ b=4\ldots\ldots\ldots(2) \end{gathered}`

So, the given equation becomes,

`f(x)=3(4^x)+2`

Next, substitute x=3, we get

`\begin{gathered} f(3)=3(4^3_{})+2 \\ =3(64)+2 \\ =192+2 \\ f(3)=194 \end{gathered}`

Hence, the solution is 194.