👏 Amazing! Helped me to learn to help my daughter. We both watched. I guess you can teach an old dog a new trick.
@thebielecmethod4884 Жыл бұрын
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!
@superdude9006 Жыл бұрын
Thanks a lot!
@thebielecmethod4884 Жыл бұрын
Anytime!
@user-gj1zm5pz2p Жыл бұрын
Sooooo helpful thank you so much!
@thebielecmethod4884 Жыл бұрын
Anytime!
@mickeryx35943 ай бұрын
Have a test tomorrow, thanks for this advice.
@thebielecmethod48843 ай бұрын
Good luck!
@neel31354 ай бұрын
thank you
@thebielecmethod48844 ай бұрын
Anytime!
@highgglsgf Жыл бұрын
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?
@thebielecmethod4884 Жыл бұрын
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.
@mohammadmahrous10 ай бұрын
what id zero term si not at the same distance?
@thebielecmethod488410 ай бұрын
Do you mind sending me the table you have an I can give you some other routes you can take
@8bitquinn6472 ай бұрын
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?
@thebielecmethod48842 ай бұрын
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.
@sara-fb9oi4 ай бұрын
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?
@thebielecmethod48844 ай бұрын
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.
@gogzog Жыл бұрын
what about if the numbers are skipping around??
@thebielecmethod4884 Жыл бұрын
Great question! There are other methods I can make a video on if you have the question your trying to answer.
@gogzog Жыл бұрын
@@thebielecmethod4884 you've helped me more than any math teacher
@stickimation78467 ай бұрын
Are the relation is the same with equation?
@thebielecmethod48847 ай бұрын
Yes if the numbers were provided to you in a relation instead of a table the result would be the same equation.
@alanakrantz4607 Жыл бұрын
what if it doesn’t have a 2nd difference?
@thebielecmethod4884 Жыл бұрын
What are your x and y values? Do you mean the 2nd difference isn’t constant?
@giljhonherbias133611 ай бұрын
@@thebielecmethod4884 yeah what if the second difference is not constant?
@thebielecmethod488411 ай бұрын
Then you aren’t dealing with a quadratic.
@Satgamer80d7 ай бұрын
quadratics always have a constant second difference, the same way lines always have a constant first difference.
@sara-fb9oi4 ай бұрын
Then its either linear or exponential not quadratic
@alymiuu10 ай бұрын
what if my x and y are both 0???
@thebielecmethod488410 ай бұрын
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.