## anonymous 5 years ago does anyone know how to find the formula for exponential functions that satisfy these given conditions? f(0)=4; F(1)=2 h(0)=100; h(1)=75 p(1)=4; p(2)=6

1. amistre64

(0,4) (1,2) we know that an exponent is normally at (0,1) so this has been modified at least by +3

2. amistre64

(0,4-3) (1,2-3) falls below the range tho :) we can try using the points to figure it out :) wrap an exponent to fit

3. amistre64

or find the inverse; with logs :)

4. amistre64

f(4) = 0 ; f(2) = 1 logb(4) = 0 ; logb(2) = 1

5. amistre64

b^1 = 2 suggests that b=2...but its modified :) so that cant be right. b^0 = 4 suggests that we are b^x + 3 y = b^x +3 y-3 = b^x 2-3 = b^(1) b = -1 ??

6. amistre64

y = -(1)^x +3 is what I seem to get :)

7. anonymous

An exponential function has the form: $y=a.b^x$ You just need to substitute using the two points given in each case, and solve for a, b. I will do the first one in the next post.

8. amistre64

yay!! someone smarter than me :)

9. amistre64

im guessing a = 4 ;)

10. anonymous

For the first one, we have the following two equations: $4=a.b^0 \rightarrow(1)$ and $2=a.b^1\rightarrow(2)$

11. anonymous

So clearly, as amistre64 just said, since b^0=1, we have a=4. Then for equation (2): $2=4b \implies b={2 \over 4}= {1 \over 2}$

12. anonymous

So we have a=4 and b=1/2, then the exponential function is: $f(x)=4({1 \over 2})^x$

13. anonymous

Does that make sense to you? and @amistre64: I am not smarter than you :)

14. anonymous

The answer does... I'm not sure I understand how you got it though

15. anonymous

Which step don't you understand?

16. anonymous

You first should know that the form of an exponential function is: $f(x)=a.b^x$ where a and b are constants.

17. anonymous

well for the next one, h would also equil to ^0 so 1 so does that make it F(x)=100x 75^1

18. anonymous

So, your goal from here is to find these two constants using the two given points in each case.

19. anonymous

?

20. anonymous

okay

21. anonymous

The formula makes sence to me, its just the way the points fit into it that confuses me

22. anonymous

I will tell you how to substitute the points. Let's take the point f(0)=4, that says when you plug x=0, you will get value f(x), h(x) (or whatever you call it) to be 4. So the number between the brackets goes for the exponent.

23. anonymous

So for h(0)= 100... H=1 because anything raised to the zero power is one then times 100 which is 100? =/

24. anonymous

Not quite, you should always think of the formula. Now, the formula looks like: $h(x)=a.b^x$ Try to use the points to find a and b.

25. anonymous

Let me show you how to use the first point to find a, and then you should try finding b. The point is h(0)=100. So, plug x=0, and 100 for h(x).

26. anonymous

100 x 1?

27. anonymous

So, doing what I just said. We have h(x)=ab^x where x=0, and h(x)=100. that gives: $h(x)=a.b^x \implies 100=a.b^0$ but b^0=1, then $100=a(1) \implies a=100$

28. anonymous

We just did about 75% of the problem. Don't forget that you're looking for values for a and b, and then plug them back in the formula. That's all. Can you use the second point to find b?

29. anonymous

would it be 75^1=75?... that seems too simple.

30. anonymous

Don't forget you want to find b, so try to plug in the formula and tell me what b would be.

31. anonymous

Take a breath :)

32. anonymous

Look closely to the formula: $h(x)=a.b^x$

33. anonymous

100 x (75)^1?

34. anonymous

We already found a=100, so we can write it as: $h(x)=100(b)^x$ RIGHT?

35. anonymous

yes

36. anonymous

So now we will use the second point h(1)=75 to find b. We should put x=1 and h(x)=75. You're close by the way :)

37. anonymous

75/100?

38. anonymous

Exactly. $75=100(b)^1 \implies 75=100b \implies b={75 \over 100}$ But you can simplify this value a little bit of you like :)

39. anonymous

Simplified b=(3/4).

40. anonymous

So, the equation would look like?!

41. anonymous

I hope I didn't complicate things to you. Anyway, all you have to do now is to substitute these to values a=100, b=3/4 into our original function, you get: $h(x)=100({3 \over 4})^x$