Question

find the best fitting quadratic model for the data given.<image>12x² + 13x + 504.3<image>0.12x² + 1.3x + 504.3<image>1.2x² + 13x + 504.3

Explanation:

Step1: Substitute \(x = 0\)

For a quadratic model \(y=ax^{2}+bx + c\), when \(x = 0\), \(y=c\). In the given data, when \(x = 0\), \(y = 489\approx504.3\) (there might be some approximation in the model selection process).

Step2: Substitute \(x = 10\)

For the model \(y = ax^{2}+bx + c\), when \(x = 10\):

• For \(y=12x^{2}+13x + 504.3\), \(y=12\times10^{2}+13\times10 + 504.3=1200 + 130+504.3=1834.3\)
• For \(y = 0.12x^{2}+1.3x + 504.3\), \(y=0.12\times10^{2}+1.3\times10 + 504.3=12+13 + 504.3=529.3\)
• For \(y=1.2x^{2}+13x + 504.3\), \(y=1.2\times10^{2}+13\times10 + 504.3=120+130 + 504.3=754.3\approx801\) (closer to the data value of \(801\) compared to the other two models at \(x = 10\))

Step3: Substitute \(x = 20\)

• For \(y=12x^{2}+13x + 504.3\), \(y=12\times20^{2}+13\times20+504.3=12\times400+260 + 504.3=4800+260+504.3=5564.3\)
• For \(y = 0.12x^{2}+1.3x + 504.3\), \(y=0.12\times20^{2}+1.3\times20+504.3=0.12\times400 + 26+504.3=48+26+504.3=678.3\)
• For \(y=1.2x^{2}+13x + 504.3\), \(y=1.2\times20^{2}+13\times20+504.3=1.2\times400+260+504.3=480+260+504.3 = 1244.3\approx1202\) (closer to the data value of \(1202\) compared to the other two models at \(x = 20\))

Answer:

\(1.2x^{2}+13x + 504.3\)