SOLUTION: Which model is most appropriate for the data set? (–1, 20), (0, 10), (1, 5), (2, 20) (1 point)linear quadratic exponential none of the above

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Question 731778:
Which model is most appropriate for the data set?
(–1, 20), (0, 10), (1, 5), (2, 20) (1 point)linear
quadratic
exponential
none of the above

Found 2 solutions by stanbon, fcabanski:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Which model is most appropriate for the data set?
(–1, 20), (0, 10), (1, 5), (2, 20) (1 point)linear
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Plot the points
Draw a smooth curve thru them
What does it look like?
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I ran a Quadratic Regression program on the data
and found the correlation factor to be 0.93.
So it is somewhat quadratic but not exactly quadratic.
Cheers,
Stan H.
==================
quadratic
exponential
none of the above

Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Plot the points.


It most closely resembles a quadratic because it goes down to a point (which would be the vertex), then it rises.


It isn't a line, because lines rise or fall.


It isn't exponential. Exponentials rise or fall towards an asymptote (a value they approach but never reach). Sometimes they have two parts with one falling towards and asymptote and one rising towards an asymptote, with the parts sharing x values where the parts rise or fall. This one would have to share x values from one part rising and the other part's approach to an asymptote.


To make sure, plug in three points to solve for the possible quadratic equation.


ax^2 + bx + c


Plug in (-1,20) a(-1)^2 + b(-1)+c=20


a-b+c=20: Equation 1


Plug in (0,10): a(0)^2 + b(0) + c = 10


c=10. That simplifies things.


Equation 1 becomes: a-b+10=20


Equation 1: a-b=10


Plug in (1,5): a(1)^2 + b(1) + 10 = 5


a+b=-5: Equation 2.


Solve equation 1 for a: a = 10+b


Substitute into eq 2: 10+b+b=-5


10+2b = -5


2b=-15


b = -7.5


a=10+b = 10-7.5 = 2.5


The quadratic equation is y = 2.5x^2 -7.5x + 10

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