Find the exponential equation whose graph passes through the points (-1, 4 3 ) and (1,12)

Question

Find the exponential equation whose graph passes through the points (-1, 4/3 ) and (1,12)

Answer 1

The problem here is that we are not given the y-intercept which would make it easy to determine the "a" value in our exponential form y=a(b)^x.

What we can do however, is "force" the coordinate (-1, 4/3) to be our temporary y-intercept as (0,4/3). We also must move the other coordinate the same number of units to reflect the change and therefore, (1,12) becomes (2,12).

We can now use 4/3 for "a" and (2,12) as our values for x and y giving us: 12=(4/3)(b)²

If we solve for b by dividing by 4/3 and taking the square root, we obtain b=3. This now gives us y=a(3)^x. (Remember, we cannot use 4/3 as "a" because we forced it to be the placeholder for the time being).

Now that we have "b", we can use one of the original points to solve for "a". Let's use (1,12).

12=a(3)¹

4=a

Therefore, our exponential equation is y=4(3)^x.