Help please? Write an exponential equation that has the points (1,6) and (-2,3/4)

Answer 1

Hello,

Lets begin by assuming the exponential function is of the form f(x) = abc and then see if we can use the points provided to find values of a and b that work.

1. Points given: (1,6) and (-2,3/4)
2. recall that any number raised to the one power is equal to itself.
3. So, f(1) = ab¹ = ab = 6
4. This means that whatever a and b are, their product will be six
5. For the other point, (-2,3/4) we have:
6. f(-2) = ab^-2 = 3/4
7. recall that a negative exponent is just a shortcut way of expressing a fraction. This means that the term ab^-2 can be expressed as a fraction as a/b²
8. This gives us f(-2) = a/b² = 3/4
9. So far we have ab = 6 and a/b² = 3/4.
10. The second expression (a/b² = 3/4) suggests that a=3 and b²=4 and so a=3 and b=2 may be a solution.
11. To test this solution note that ab = (3)(2) = 6 which checks out.
12. The sought after equation then is:
13. f(x) = 3(2^x)
14. f(1) = 3(2¹) = 3(2) = 6, and
15. f(-2) = 3(2^-2) = 3/2² = 3/4
16. The given points are in fact included in our equation.

I hope this helps with this problem, but also that the process will help you with other related problems.

All the best,

David

Answer 2

You are solving a system of equations y₁ = ab^x1 and y₂ = ab^x2, so plug the x and y coordinates into the equations to solve for a and b:

6 = ab¹ & 3/4 = ab^-2

Since a = 6/b, we can use this in the second equation to solve for b:

3/4 = (6/b)b^-2

3/4 = 6/b³

6/(3/4) = b³

8 = b³

2 = b,

which means that a = 3.

Now we just input our a and b values into y = ab^x, so our equation is y = 3*2^x.