One technique that you can apply when solving such a problem is trial and error. We try to use each equation to prove that a given value of <em>x</em> on the table given will correspond to the value of <em>y</em> on the table.
a) Let's try to put x = 3 for the first equation and we must get an answer equal to 2.25.
![y=7.72(3)-29.02=-5.86_{}](https://tex.z-dn.net/?f=y%3D7.72%283%29-29.02%3D-5.86_%7B%7D)
Since the value of <em>y</em> is not equal to 2.25 and the deviation is too large. this equation is not a good model,
b) We put x = 3 on the second equation and solve for <em>y</em>
![y=-7.52(3)^2+0.19(3)+3.26=-63.85](https://tex.z-dn.net/?f=y%3D-7.52%283%29%5E2%2B0.19%283%29%2B3.26%3D-63.85)
Since the value of <em>y</em> is not equal to 2.25 and the deviation is too large. this equation is not a good model,
c) We put <em>x</em> = 3 on the third equation and solve for <em>y,</em>
![y=0.4(3)^2+0.79(3)-4.93=1.04](https://tex.z-dn.net/?f=y%3D0.4%283%29%5E2%2B0.79%283%29-4.93%3D1.04)
Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of <em>x</em>. If x = 5, we get
![y=0.4(5)^2+0.79(5)-4.93=9.02](https://tex.z-dn.net/?f=y%3D0.4%285%29%5E2%2B0.79%285%29-4.93%3D9.02)
which has a slight deviation on the given value of <em>y</em> on the table for <em>x</em> = 5. let's try for <em>x</em> = 7. We have
![y=0.4(7)^2+0.79(7)-4.93=20.2](https://tex.z-dn.net/?f=y%3D0.4%287%29%5E2%2B0.79%287%29-4.93%3D20.2)
and the answer has a small deviation compared to the actual value given. The other values of <em>x</em> can again be put on the equation and check their corresponding value of <em>y</em>, and the resulting values are as follows
![\begin{gathered} y=0.4(8)^2+0.79(8)-4.93=26.99 \\ y=0.4(12)^2+0.79(12)-4.93=62.15 \\ y=0.4(14)^2+0.79(14)-4.93=84.53 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D0.4%288%29%5E2%2B0.79%288%29-4.93%3D26.99%20%5C%5C%20y%3D0.4%2812%29%5E2%2B0.79%2812%29-4.93%3D62.15%20%5C%5C%20y%3D0.4%2814%29%5E2%2B0.79%2814%29-4.93%3D84.53%20%5Cend%7Bgathered%7D)
And as you can see, the deviation of values from the table to calculated becomes smaller. Hence, this is the best model.
d) We put <em>x</em> = 3 on the third equation and solve for <em>y,</em>
![y=4.19(1.02)^3=4.45_{}_{}](https://tex.z-dn.net/?f=y%3D4.19%281.02%29%5E3%3D4.45_%7B%7D_%7B%7D)
Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of <em>x</em>. If x = 5, we get
![y=4.19(1.02)^5=4.63](https://tex.z-dn.net/?f=y%3D4.19%281.02%29%5E5%3D4.63)
where the answer's deviation is too large compared to the value of <em>y</em> if x = 5 on the table given.
Based on the calculations used above, the best equation that can be a good model is equation 3.