Answer-
The best-fit quadratic equation is [tex]y=14.9786x^2+3.1057x+56.2771[/tex] and the game cost in 2010 will be $1586
Solution-
Plotting a table taking year as input variable and cost as output variable.
X= year - 2000
Y= cost in dollar.
Quadratic equation formula,
[tex]y=ax^2+bx+c[/tex]
[tex]a=\frac{(\sum x^2y\sum xx)-(\sum xy\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}[/tex]
[tex]b=\frac{(\sum xy\sum x^2x^2)-(\sum x^2y\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}[/tex]
[tex]c=\frac{\sum y}{n}-b\frac{\sum x}{n}-a\frac{\sum x^2}{n}[/tex]
Where,
[tex]\sum xx=\sum x^2-\frac{(\sum x)^2}{n}[/tex]
[tex]\sum xy=\sum xy-\frac{\sum x\sum y}{n}[/tex]
[tex]\sum xx^2=\sum x^3-\frac{\sum x\sum x^2}{n}[/tex]
[tex]\sum x^2y=\sum x^2y-\frac{\sum x^2\sum y}{n}[/tex]
[tex]\sum x^2x^2=\sum x^4-\frac{(\sum x^2)^2}{n}[/tex]
Putting the values, we get
[tex]a=14.9786,b=3.1057,c=56.2771[/tex]
Putting these in the quadratic equation,
[tex]y=14.9786x^2+3.1057x+56.2771[/tex]
In order to get the cost of game in 2010, we can put x=10 to get the value of y or the cost of game.
[tex]y(10)=14.9786(10)^2+3.1057(10)+56.2771=1585.1941 \approx 1586[/tex]
