Question

Write the equation of an exponential function that passes through the points

(1,12) and (3,108)

Answer 1

y=4(3^x).

Explanation:

To find the equation of an exponential function passing through the given points (1,12) and (3,108),

we can use the standard exponential form y=a(b^x). We know that when x=1, y=12,

so we can substitute these values into the equation to find a.

So 12 = a(b^1). Similarly, when x=3, y=108, so 108 = a(b^3). We can divide the second equation by the first to eliminate a and get (108/12) = b^2, or 9 = b^2. Thus, b=3 (taking only the positive root). We can now substitute this value of b into either equation to find a. Using the first equation, we get 12 = a(3^1), so a=4. Therefore, the exponential function passing through the given points is y=4(3^x).

Answer 2

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

Explanation:

The general equation for an exponential function is:

`\boxed{f(x) = ab^x}`

where:

• a is the initial value or y-intercept.
• b is the base or growth factor.

To find the values of a and b that satisfy the given conditions, we can use the two points (1, 12) and (3, 108) to form a system of equations:

`\begin{cases}12 = ab^1\\108 = ab^3\end{cases}`

Divide the second equation by the first equation to eliminate a:

`(ab^3)/(ab) = (108)/(12)`

`b^2=9`

`√(b^2)=√(9)`

`b=3`

Substitute the found value of b into the first equation and solve for a:

`12&=3a`

`(12)/(3)=(3a)/(3)`

`4=a`

`a=4`

Therefore, the equation of the exponential function that passes through the points (1, 12) and (3, 108) is:

`\boxed{f(x) = 4(3)^x}`