Which expression can be placed on the right side of the equals sign to form an equation. 2+33=
1 answer:
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Answer:
D. y = 12.11·1.36^x . . . . . best <em>exponential</em> model
A. y = 2.49x^2 +7.29x +3.57 . . . . . best model of the choices offered
Step-by-step explanation:
None of the offered choices is very good, and the best of them is only the best by a relatively small amount. The exponential models cannot have been developed using any reasonable approach to model building. So, you have to figure out how to assess these models when none of them has much relation to the given data.
The horizontal asymptote of an exponential function is zero. Here, the values seem to have a horizontal asymptote of about 6. Consequently, you would look at the first few numbers and expect a vertical offset to the exponential function of about 6. Subtracting that from the remaining numbers, you might toss "13" as an outlier and use some of the others to compute the base of the exponent. If you do that using the points (0, 8) and (4, 88), you would compute the base to be ...
((88 -6)/(8 -6))^(1/(4-0)) ≈ 2.53
The multiplier of the exponential part is its value when x=0. We've estimated that to be 8-6 = 2. This gives an exponential model that looks like ...
y = 6 + 2·2.53^x
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<em>Comparison of our estimated model to the offered choices</em>
Selection A is a <em>quadratic</em> model. It has precisely the coefficients that are calculated by a spreadsheet or graphing calculator for a quadratic model. Though it gives the best fit of any of the offered choices, it is not the required <em>exponential</em> regression equation.
The base of selection C is about 5 times as great as needed, so it will vastly overestimate any points for x ≥ 1. It gives by far the worst fit of all of these choices.
The choice with the largest reasonable base is B, but it gives values for y that are less than any of those listed. As a consequence, its error is higher than necessary.
The remaining choice D gives a curve that is less than data points at the ends of the table and greater than the data values in the middle of the table. Of the offered exponential models, it has the least overall error. (You might pick choice D simply because it uses the same numbers as in choice C, but puts those numbers in places that make the function a better fit.)
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<em>Graphing calculator results</em>
If you ask a graphing calculator to give you an exponential model for these data, you can get either of ...
- y = 6.2 + 1.2·2.87^x . . . . . exponential with vertical offset
- y = 3.38·2.25^x . . . . no vertical offset; can't match x<0 very well
The first of these has about 1/10 the error of the last of these. Both have less than 1/10 the error of the available answer choices.
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<em>Comment on the attachment</em>
The black and red curves correspond to the first two answer choices. The functions associated with the other colors are shown at the left. The green points are those given in the problem statement.
The "total" numbers are the total squared error of each of the functions. Smaller errors mean the function is a better fit to the data.
Put the equation into the form y = ( x − h )^2 + k
Given equation is
We apply completing the square method
we take coefficient of middle term and then divide by 2 and then square it
coefficient of middle term is 26
Divide by 2, it becomes 13
then we square it (13)^2 = 169
Add and subtract 169
We got the equation in vertex form