## anonymous one year ago A punter kicks a football upward with an initial speed of 48 feet per second. After how many seconds does the ball hit the ground? Use the formula h = rt − 16t2 where h represents height in feet and r represents the initial speed.

1. anonymous

So it looks like this? $h = 48t-16t^2$

2. anonymous

yea

3. anonymous

so now what?

4. anonymous

Graph it

5. anonymous

cant graph that

6. anonymous

7. campbell_st

so you need to le the height = 0 and then solve 0 = 48t - 16t^2 for t this can be factored

8. anonymous

Now try to put it into quadratic form like $48t-16t^2-36=0$

9. anonymous

Then factor this and you should get 2 values.

10. anonymous

4(8t-9)

11. anonymous

12. anonymous

t=3,0

13. anonymous

yes

14. anonymous

thats the amswer?

15. anonymous

so the answer is 3 seconds

16. anonymous

thx

17. anonymous

A company launches 4 new products. The market price, in dollars, of the 4 products after a different number of years, x, is shown in the following table: Product Function Year 1 (dollars) Year 2 (dollars) Year 3 (dollars) Product 1 f(x) = x3 1 8 27 Product 2 g(x) = x2 + 11 12 15 20 Product 3 h(x) = 4x 4 16 64 Product 4 j(x) = 4x + 8 12 16 20 Based on the data in the table, for which product does the price eventually exceed all others? Product 1 Product 2 Product 3 Product 4

18. anonymous

can u help with this

19. anonymous

h(x) is this? $4x^4$

20. anonymous

k

21. anonymous

If it is product 3 will exceed all other price, it's increasing exponentially

22. anonymous

so c

23. anonymous

yes

24. anonymous

wait

25. anonymous

It's product 1, product 3 is only 4x

26. anonymous

kk

27. anonymous

A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(t), in feet, at different times, t, in seconds: f(t) = −16t2 + 48t + 100 The average rate of change of f(t) from t = 3 seconds to t = 5 seconds is _____feet per second.

28. anonymous

$\left[\begin{matrix}Product & Function & Year 1 & Year 2 & Year 3\\ 1 & f(x)=x^3 & 1 &8 &27 \\ 2 & g(x) = x^2+11 & 12 &15&20 \\ 3 & h(x) = 4x & 4&16& 64 \\4 &j(x)=4x+8 &12 &16 &20\end{matrix}\right]$

29. anonymous

that should seem clearer when i put it this way

30. anonymous

Ok take the derivative of the function f(t)

31. anonymous

I hope you know how to do that

32. anonymous

can i just solve for t

33. anonymous

no you can't solve for t because every value of t gives you something new, its a function

34. anonymous

and you are finding the rate of change

35. anonymous

This curve looks like this|dw:1438220902622:dw|

36. anonymous

so you are solving for the tangent line like this|dw:1438221106783:dw|

37. anonymous

omg

38. anonymous

@saseal

39. anonymous

yea?

40. anonymous

this is hard

41. anonymous

Ok...I'll show you the formula first, you try it then I show to the other way $f'(x)= \frac{ f(x+h)-f(x) }{ h }$ h is 5-3=2

42. anonymous

so you are basically doing $f'(x)=\frac{ f(5)-f(3) }{ 2 }$

43. anonymous

kk

44. anonymous

how would i solve that

45. anonymous

@saseal

46. anonymous

f(5) is $f(5)=-16(5)^2+48(5)+100$

47. anonymous

48. anonymous

no this is just f(5)

49. anonymous

remember you need to take f(5)-f(3) and divide by 5-3

50. anonymous

so divide 5 and 3

51. anonymous

umm no

52. anonymous

guess ill show you the short way since i have to go soon its some calculus magic but it works

53. anonymous

ok

54. anonymous

It's -80

55. anonymous

thx