Question

How would I figure this out?
Find a quadratic function to model the values in the table. Predict the value of y for x = 6.

Answer 1

I've never done these before. So i really need help!

Answer 2

Well it sounds like you need to use quadratic interpolation which is a long winded explanation. This site lays out the process so take a look at it and see if that doesn't help first. http://isezen.com/2012/01/15/quadratic-interpolation-three-point/

Answer 3

Umm. i don't really understand.

Answer 4

Well basically you will have 3 equations with 3 unknowns. Set up goes like this:
Here is the template equation:\[y = a0 + a1*x + a2*x^2\]Now the three unknowns you solve for are a0, a1, and a2. To do this you use the given data and plug it in for the three data points.
These are the three equations:
\[2 = a0 + a1*(1) + a2*(1)^2\]\[2 = a0 + a1*(0) + a2*(0)^2\]\[-10 = a0 + a1*(3) + a2*(3)^2\]Three equations and three unknowns can be solved by hand or using matrix algebra.

Answer 5

Does that help? @YanaSidlinskiy

Answer 6

Oh, Ok. Yea. It starts to make a little sense now. Do you mind continuing?

Answer 7

the data provides a simpler way

Answer 8

since there are 2 x values with the same y value, the axis of symmetry has to be the middle between them

Answer 9

Well no I don't mind but that's all there is to it really. Once you get a0, a1 and a2 you will use the template equations, plug in the values, and you will have f(x) = y = #x^2 + #x + #. If you want to evaluate at x = 6, f(6) = the predicted value as asked.

Answer 10

since x=0 and y=2, we know the y intercept is 2

Answer 11

Yes there's usually a 'simpler' way but this way will always work without fail.

Answer 12

yeah, it will work

Answer 13

So would the equation be like..\[-2x^2+2x-2; -58\]
??

Answer 14

-2 on the start looks fine, but the end should be +2, not sure about the middle yet

Answer 15

yeah, middles a +2x works well

Answer 16

Ok. So are you sure that the end should be +2. Correct?

Answer 17

well, when x=0, y=2 ... not -2 so yeah

Answer 18

Here's my answer choices:
A. y = -2x2 + 2x -2; –58
B. y = 2x2 – 2x -2; 60
C. y = 2x2 – 2x -2; 58
D. y = -2x2 + 2x +2; –58

Answer 19

only one of them fits for the point (0,2)

Answer 20

unless your table data is wrong :/

Answer 21

Nope. None of it is wrong:D

Answer 22

Its the last one. Right?

Answer 23

Good job, yes the last one is correct. Just to revisit the other method for future use. If you did interpolation, you'd find a0, a1, and a2 came out to be 2, 2, and -2 respectively. Thus the equation is y = - 2*x^2 + 2*x + 2
evaluating at 6 gives
y(6) = -58

Answer 24

Thank you sooo much!! @lasttccasey and @amistre64 I appreciate your help!!:D I'm definitely writing this down in my notebook for Algebra 2!!:D

Answer 25

good luck :)

Answer 26

Thanx:D