## heyitslizzy13 one year ago can someone help me with a problem please!!

1. heyitslizzy13

2. heyitslizzy13

@satellite73 @Directrix @jim_thompson5910

3. anonymous

which part of the problem?

4. heyitslizzy13

The first part for now :)

5. freckles

Hint: Intermediate Value theorem

6. heyitslizzy13

@freckles do you know any good websites to review the theorem?

7. anonymous

$f(x) = ax^3-11x^2+bx+c \qquad f(-3)=10, \qquad f(0)=4,\qquad f(-1)=4$$f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10$$f(-3)=-27a-99-3b+c=10$$f(-3) = -27a-3b+c=109$ ...perhaps we can solve this as a system of equations?

8. heyitslizzy13

could you show me how? please and thank you :)

9. jim_thompson5910

part a) imagine you have a river that is a complete straight line |dw:1443850668196:dw| this straight line would stretch forever in both directions

10. jim_thompson5910

now let's say we had point A and point B one point on either side |dw:1443850712927:dw|

11. jim_thompson5910

if we connect the two points with a continuous function, then it is impossible to not pass through the line. we might have something like this |dw:1443850761686:dw| the path has to cross over the river somewhere

12. heyitslizzy13

okay

13. jim_thompson5910

now imagine the river is the x axis if point A has a positive y coordinate (ie it's above the x axis) and point B has a negative y coordinate (ie it's below the x axis) then the continuous function must cross the x axis at some point between A and B. So if we know A is above the x axis and B is below, there is at least one root in between

14. anonymous

$f(0) = a(0)^3 -11(0)+b(0)+c =4$$f(0) = c=4$ $f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4$$f(-1) = -a-11-b+c=4$$f(-1) =-a-b+c=15$ $\therefore$$f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15$ As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be $$c=4$$ if i'm not mistaken.

15. heyitslizzy13

I am kind of confused, how do we know that c=4 is the root that lies in the middle?

16. anonymous

because when $$x=0$$ (that means it's at the origin) $$y=4$$

17. heyitslizzy13

ohhh okay

18. jim_thompson5910

Jhannybean plugged in x = 0 and everything went away basically except for that last term

19. anonymous

Exactly

20. heyitslizzy13

do you understand what part a of the problem is asking? I'm quite confused

21. anonymous

Okay so between $$x=-3$$ and $$x=0$$, is there ANYWHERE on the x-axis when the function gives you a solid number? hint: look at the above posts and our explanations.

22. jim_thompson5910

if you need a visual way to approach this, draw a blank xy axis |dw:1443851396111:dw|

23. heyitslizzy13

at -1 i believe , (-1, 4) ?

24. jim_thompson5910

then plot the point (-3,10), from the fact that f(-3) = 10 and the point (0,4), because f(0) = 4 |dw:1443851450292:dw|

25. jim_thompson5910

now, ask yourself this: is it possible to connect the two points without crossing the x axis?

26. heyitslizzy13

it is possible

27. jim_thompson5910

so there is no guarantee we have a root on the interval from x = -3 to x = 0

28. jim_thompson5910

now IF we had (0,-4) instead, then we shift down the second point |dw:1443851620081:dw|

29. jim_thompson5910

|dw:1443851638079:dw|

30. jim_thompson5910

now it's impossible to connect the two points without crossing the x axis so there has to be at least one root between the two x values |dw:1443851682979:dw| or |dw:1443851692524:dw| or |dw:1443851699567:dw|

31. heyitslizzy13

ohhh okay!

32. jim_thompson5910

so really the only guarantee that we have a root is if there is a sign change (in my example the sign change goes from y = 10 to y = -4)

33. jim_thompson5910

sign change in the y values or function outputs I mean

34. jim_thompson5910

but since y = 10 to y = 4 has no sign changes in the problem, there is no guarantee there's a root between x = -3 to x = 0

35. heyitslizzy13

oh i see now

36. heyitslizzy13

could you help me on part b?

37. jim_thompson5910

ok so I think you found that c = 4 right?

38. jim_thompson5910

do you see how Jhannybean got this? $f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15$

39. heyitslizzy13

yes :)

40. heyitslizzy13

wait i don't understand how she got the 109, 4 and15

41. jim_thompson5910

I think f(-3) should have been 10 and not 109

42. heyitslizzy13

okay, and how is f(-1)= 15?

43. jim_thompson5910

it should be $f(-3) = -27a-3b+c=10\\ f(0) =c=4 \\f(-1)=-a-b+c=4$

44. heyitslizzy13

oh okay, i understand then

45. jim_thompson5910

oh wait, there's an 11 in there that I missed, let me rethink

46. jim_thompson5910

Jhannybean has it right, I missed a term

47. anonymous

$f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10$$f(-3) = -27a -99 -3b +c = 10$$f(-3) = -27a -99 -3b + c +99 = 10 + 99$$f(-3) = -27a -3b+c=109$

48. anonymous

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @Jhannybean $f(0) = a(0)^3 -11(0)+b(0)+c =4$$f(0) = 0+0+0+c=4 \\ f(0)=c=4$ $f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4$$f(-1) = -a-11-b+c+11=4+11$$f(-1) =-a-b+c=15$ $\therefore$$f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15$ As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be $$c=4$$ if i'm not mistaken. $$\color{#0cbb34}{\text{End of Quote}}$$

49. heyitslizzy13

oh ok

50. anonymous

Does that make more sense, @heyitslizzy13 ?

51. heyitslizzy13

yes :)

52. anonymous

Great $$\checkmark$$

53. jim_thompson5910

that root would be c=4  I don't know about that part

54. heyitslizzy13

so we know c is 4, what about a and b?

55. anonymous

Haha, I meant $$x=0$$ x_x

56. jim_thompson5910

ah I gotcha

57. heyitslizzy13

would it be 109 and 15?

58. jim_thompson5910

since c = 4 we can replace each 'c' with 4 $-27a-3b+\color{red}{c}=109\\-a-b+\color{red}{c}=15$ $-27a-3b+\color{red}{4}=109\\-a-b+\color{red}{4}=15$

59. anonymous

So then you would solve it either by substitution, or using the elimination method :)

60. heyitslizzy13

can you walk me through either one please? I know what they are but I'm a little lost rn

61. anonymous

First we want to isolate a and b in both equations to one side, therefore we need to start by subtracting -4 to both sides of both the equations. $-27a-3b+4-4=109 -4$$-a-b+4-4=15-4$ Simplifying both of these, we get: $-27a-3b=105$$-a-b=11$ With me so far?

62. heyitslizzy13

yes!

63. anonymous

So I personally like using the elimination method. How that works is we're trying to eliminate one of the variables by using a multiplier so when we add the two equations together, we'll be left with one variable, then we can use that to solve for the other. Here, i'm going to eliminate $$b$$. in order to eliminate $$b$$, i need to multiply the second equation by $$-3$$. $-27a-3b=105$$\color{red}{-3}(-a-b=11)$ $-27a-3b=105$$\color{red}{+3}a\color{red}{+3}b=\color{red}{-33}$ Adding these two equations together, we will get $(-27+3)a +(-3+3)b =105-33$$-24a+0b=72$Now we just solve for $$a$$.

64. anonymous

Is this making sense, @heyitslizzy13 ?

65. heyitslizzy13

yes!! :)) is a=-3?

66. anonymous

Yes it is :)

67. anonymous

Now we just plug $$a=-3$$ back into one of the two equations and we can solve for $$b$$ I'm going to plug it into $$-a-b=11$$ . $-(-3)-b=11$$3-b=11$$-3+3-b=11-3$$-b=8$$b=-8$

68. heyitslizzy13

okay thanks :)))

69. heyitslizzy13

can you help me on c also?

70. jim_thompson5910

for part c) I would graph the function with a graphing calculator if you don't have one, use something like desmos https://www.desmos.com/calculator

71. heyitslizzy13

thanks!!

72. jim_thompson5910

the cool thing about desmos, as compared to a TI83, is that you can use sliders to change the graph https://www.desmos.com/calculator/tvrsgalven

73. heyitslizzy13

thank you both so much!!!! :)))