## anonymous one year ago Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work

1. anonymous

@UsukiDoll

2. perl

You need a system of equations

3. UsukiDoll

I don't feel well today. I woke up with a migraine (again)

4. anonymous

Take some advil and drink water

5. perl

$$\Large{ y = ax^2 + bx + c \\~\\ -20 = a(-2)^2 + b(-2) + c \\-4 = a(0)^2 + b(0) + c \\ -20 = a(4)^2 + b(4) + c }$$

6. anonymous

i honestly have no clue how to do this equation

7. perl

Do you agree with what I have so far. I plugged in the values into a generic quadratic function.

8. anonymous

I think so, i wouldn't know if you were wrong

9. perl

Finding a quadratic model means fitting the points to a quadratic curve $$\large y = ax^2 + bx + c$$ . So our job is to find the parameters $$\large a,b,c$$

10. anonymous

@perl how do i find the a,b,c

11. perl

We have these three equations so far. $\\~\\\Large { -20 = a(-2)^2 + b(-2) + c \\-4 = a(0)^2 + b(0) + c \\ -20 = a(4)^2 + b(4) + c }$

12. perl

the second equation gives us -4 = 0 + 0 + c -4 = c

13. anonymous

So c=-4

14. perl

Yes. Now back substitute.

15. perl

We have these three equations so far. $\\~\\\Large { -20 = a(-2)^2 + b(-2) + (-4) \\-4 = a(0)^2 + b(0) + (-4) \\ -20 = a(4)^2 + b(4) + (-4) }$

16. perl

we don't need the second equation any longer

17. perl

$\\~\\\Large { -20 = a(-2)^2 + b(-2) + (-4) \\ -20 = a(4)^2 + b(4) + (-4) }$

18. anonymous

im writing as we go

19. perl

$\\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b }$

20. anonymous

21. perl

We can solve for b

22. anonymous

So we haven't solved A or C yet?

23. perl

we solved for c = -4

24. anonymous

Ok so how do we solve for A and B than? Im sorry i really dont understand

25. anonymous

@perl

26. perl

we can solve that linear equation

27. anonymous

how?

28. perl

$\\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = 16a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = 16a -2b \\ -16 = 16a + 4b \\~\\ \rm add ~ the ~ two ~ equations\\~\\ \\-32 = 2b \\~\\ \\b = \frac{-32}{2} }$

29. anonymous

So b=-16?

30. perl

yes

31. anonymous

So now we need to find A

32. perl

y = -2x^2 + 4x -4 should be the answer. I have an error somewhere above.

33. anonymous

so b should really be b=4?

34. anonymous

A=-2 B=4 C=-4?

35. perl

yes

36. perl

I have to go back and find the error now ;)

37. anonymous

Awesome thank you so much!

38. perl

but this method is sound if you want to try it again

39. perl

if you ever*

40. anonymous

yeah i prob will, i have a whole other algebra class ahead of me

41. perl

$\\~\\\Large { -20 = a(-2)^2 + b(-2) -4 \\ -20 = a(4)^2 + b(4) -4 \\~\\} \rm simplify \\~\\\Large { -20 = \color{red}4a -2b -4 \\ -20 = 16a + 4b -4 \\~\\ \rm simplify ~ more\\~\\ -16 = \color{red}4a -2b \\ -16 = 16a + 4b \\~\\ \rm solve ~this~ system~ of~ equations\\~\\ \\ a = -2 ~ , b = 4 \\~\\ \rm backsubstitute \\y = -2x^2 + 4x - 4 }$

42. perl

Another way to solve this problem is to note that your parabola has a y intercept of -4, because x=0 at the y intercept. That reduces the problem of solving a linear system of 2 equations as we have above.