Question

Medal For HELP <3
What kind of function best models the set of data points (-3,18), (-2,6),(-1,2),(0,11), and (1,27) ?

A Linear
B Quadratic
C Exponential
D none of the above

Answer 1

@RaeDeSol :) After 2 more and Im Freeeee

Answer 2

First step would be to make a plot. Have you done so?

Answer 3

you try to do the slope thing that i did for the last one. you saw my work..... give it a shot to see if its linear

Answer 4

B Quadratic

Answer 5

Do you know about arithmetic and geometric sequences? If you do, and the data set you are given is sampled at regular intervals like this one (x = -3, -2, -1, 0, 1) you can check to see if you have an arithmetic sequence (in which case it is linear) or geometric sequence (in which case exponential)

Answer 6

If it is quadratic, you must be able to fit a quadratic polynomial to the points.

Answer 7

yes.



wanna know a short cut?

Linear? no because your y gets small and then big
Quadratic? yes because y gets small and then big
Exponential? no because it doesnt continue to get very big or very small

Answer 8

Care to show the quadratic polynomial that fits all of these points? :-)

Answer 9

A quadratic polynomial will be of the form \[ax^2+bx+c\]

We have 5 points, so that's more than enough to determine 3 coefficients.

We plug in 3 known points to get 3 simultaneous equations in \(a,b,c\), then solve for the values of \(a,b,c\).

For the first point:
\[18 = a(-3)^2+b(-3)+c\]\[18 = 9a-3b+c\]

second point:
\[6 = a(-2)^2 + b(-2) + c\]\[6 = 4a-2b+c\]

third point:
\[2 = a(-1)^2+b(-1)+c\]\[2 = a-b+c\]

Solving the system, we get \[a=4,\,b=8,\,c=6\]or \[4x^2+8x+6\]

Answer 10

Here's a plot of our data points, connected by straight lines:

Answer 11

Here's a plot of the parabola \(y = 4x^2+8x+6\)

Answer 12

Here's a plot of both the data points and the parabola I constructed. Notice that the first three points fall on the parabola (good!) but the others do not (bad!)

Answer 13

as long as you can identify the end behaviors of a graph, you can determine the type of graph. knowing how to find the coefficients is a good skill too

Answer 14

A quadratic (aka a parabola) is always symmetrical about the vertex. There isn't any point to put the vertex in this data set where it will be symmetrical.

Answer 15

No, determining the end behavior is not sufficient to determine the type of graph.

Answer 16

ok Linear

Answer 17

No, a linear graph would be a straight line...

Answer 18

ugg

Answer 19

If "best models" means that the model must coincide with the actual data points we know, then the correct answer is "none of the above".

Answer 20

Wow i'm sweating ty

Answer 21

However, as we have 5 points, we can fit a polynomial of the form \[y = ax^4 + bx^3 + cx^2 + dx + e\]that will exactly match the data. I happen to have such a polynomial right here :-)

\[y=-\frac{11 x^4}{24}-\frac{23 x^3}{12}+\frac{95 x^2}{24}+\frac{173 x}{12}+11\]

If I plot it, and overlay it with the original graph, this is what I get:

Answer 22

There's an x at each of our original data points, and you can see the two coincide at those points. I constructed this ugly polynomial in the same fashion as I did the first polynomial, except of course I had 5 equations in 5 unknowns instead of 3 equations in 3 unknowns.

Answer 23

but then its no longer a quadratic.... the exponents are too high, right? or am i confusing terms?

Answer 24

and the leading negative.... doesnt that make it an upside down you?

Answer 25

Yes, that's exactly right — it is not a quadratic. It has the same behavior as our data set over that small domain, but the bigger picture is much different!

Answer 26

that means its actually a quartic....?

Answer 27

So, this is an example of how the end behavior (which is going to negative infinity on both ends) is not necessarily a valid predictor of what you've got...

Yes, the polynomial I gave you is a quartic.

Answer 28

So you can fit polynomials to match just about anything, if you're willing to do enough work, and the approximation doesn't have to be valid from -infinity to +infinity.

Answer 29

the end behaviors aren't wrong.... its just that this data set doesn't display them.
so then this problem is tricky....

Answer 30

Yes, you can get the end behaviors from the equation type, but not necessarily the other way around.

Answer 31

Ooo Lala

Answer 32

so then for this data set, its not a quadratic. its a quatric. so none of the above

Answer 33

Yes, that is correct. If we interpret "best models" to mean that the model must generate the data points, none of the choices will work. If we are more relaxed and just ask that one of them have the same approximate shape, then a quadratic could work as an approximation.

Answer 34

This is a tricky question... If we had a better definition then that would solve the problem