Haha thank you so much for this comment. This is exactly why I make the videos I do. Love hearing you and your daughter are learning together!!! Tell her good luck in class and continue to be the great dad you are!
hello, I've been struggling with this one x | y 0.01 0.06 > 0.18 0.02 0.24 > 0.12 > 0.3 0.03 0.54 > 0.12 > 0.43 0.04 0.96 also when x = 0, y = 0 using your reverse method. 2nd difference of this table is a = 0.12 ÷ 2 = 0.06 y = ax² + bx + c and using algebra to solve for b gives me different b values. b1 = 5.99 (0.01, 0.06) b2 = 11.9988 (0.02, 0.24) b3 = 17.9982 (0.03, 0.54) b4 = 23.99 (0.04, 0.96) thus, i cant pinpoint the precise quadratic formula on hand so i used an online quadratic regression calculator and got this formula y = 600(x)² + 1.0516e^-14(x) + -9.7145e^-17 Another way is to solve for a in y = ax2, which gives the expression y = 600(x)² - the simplest formula yet Does this imply that this procedure cannot be applied to some other table of values? would you please explain this one?
Yes precisely. This method becomes more difficult when dealing with more complex numbers as your dealing with here. I can make another video showing you a different route for this particular problem when I get back from spring break.
Quick question, lets say the first points on the table were (5,36), then how would you find c because it would be difficult to keep going back to 5,4,3,2,1,&0th term? Is there a formula or any way to find it out?
Yea, you can write 3 equations using y=ax^2+bx+c and replacing the x and y. Your example of (5,36) would be 36=a(5)^2+b(5)+c 36=25a+5b+c Use your other 2 points to make 2 more equations. Then solve the system of equations for a, b, c. This is another way that can be done fairly quickly.
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👏 Amazing! Helped me to learn to help my daughter. We both watched. I guess you can teach an old dog a new trick.
Haha thank you so much for this comment. This is exactly why I make the videos I do. Love hearing you and your daughter are learning together!!! Tell her good luck in class and continue to be the great dad you are!
this helped me so much! i have a geometry test next period and now i’m confident about it!!
Love to hear it! Good luck
Have a test tomorrow, thanks for this advice.
Good luck!
what about if the numbers are skipping around??
Great question! There are other methods I can make a video on if you have the question your trying to answer.
@@thebielecmethod4884 you've helped me more than any math teacher
hello, I've been struggling with this one
x | y
0.01 0.06
> 0.18
0.02 0.24 > 0.12
> 0.3
0.03 0.54 > 0.12
> 0.43
0.04 0.96
also when x = 0, y = 0 using your reverse method.
2nd difference of this table is a = 0.12 ÷ 2 = 0.06
y = ax² + bx + c
and using algebra to solve for b gives me different b values.
b1 = 5.99 (0.01, 0.06)
b2 = 11.9988 (0.02, 0.24)
b3 = 17.9982 (0.03, 0.54)
b4 = 23.99 (0.04, 0.96)
thus, i cant pinpoint the precise quadratic formula on hand
so i used an online quadratic regression calculator and got this formula
y = 600(x)² + 1.0516e^-14(x) + -9.7145e^-17
Another way is to solve for a in y = ax2, which gives the expression
y = 600(x)² - the simplest formula yet
Does this imply that this procedure cannot be applied to some other table of values?
would you please explain this one?
Yes precisely. This method becomes more difficult when dealing with more complex numbers as your dealing with here. I can make another video showing you a different route for this particular problem when I get back from spring break.
what id zero term si not at the same distance?
Do you mind sending me the table you have an I can give you some other routes you can take
Thank you so much. Can u make a video making a vertex equation from a table of values?
Sure can. When do you need it by? I have a few other videos I’m making this next week and then I can add that one to my list.
Are the relation is the same with equation?
Yes if the numbers were provided to you in a relation instead of a table the result would be the same equation.
What happens if my y-intercept numbers are irregular? For example it goes down -3, and then -3 again, and then -9, and then -15?
Could you send me a list of x and y values to look at as an example? Having -3 twice shouldn’t happen and I’d have to see the x values.
Quick question, lets say the first points on the table were (5,36), then how would you find c because it would be difficult to keep going back to 5,4,3,2,1,&0th term? Is there a formula or any way to find it out?
Yea, you can write 3 equations using y=ax^2+bx+c and replacing the x and y. Your example of (5,36) would be 36=a(5)^2+b(5)+c
36=25a+5b+c
Use your other 2 points to make 2 more equations. Then solve the system of equations for a, b, c. This is another way that can be done fairly quickly.
Thanks a lot!
Anytime!
Sooooo helpful thank you so much!
Anytime!
what if it doesn’t have a 2nd difference?
What are your x and y values? Do you mean the 2nd difference isn’t constant?
@@thebielecmethod4884 yeah what if the second difference is not constant?
Then you aren’t dealing with a quadratic.
quadratics always have a constant second difference, the same way lines always have a constant first difference.
Then its either linear or exponential not quadratic
what if my x and y are both 0???
Then your c term would be 0. This equation would be written out as y=ax^2+bx, with no value written down for c.
@@thebielecmethod4884thank u!!
thank you
Anytime!