Question

Year (X) =2000 2001 2002 2003 2004

Answer 1

Here we are going to use the equation y=15x2+3x+56.
x=0 in 2,000 so x=10 in 2010. 
Substitute 10 for x.
After this we have the answer: D. $1,586

Answer 2

Answer-

The best-fit quadratic equation is  $y=14.9786x^2+3.1057x+56.2771$ 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,

$y=ax^2+bx+c$

$a=\frac{(\sum x^2y\sum xx)-(\sum xy\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}$

$b=\frac{(\sum xy\sum x^2x^2)-(\sum x^2y\sum xx^2)}{(\sum xx\sum x^2x^2)-({\sum xx^2)}^2}$

$c=\frac{\sum y}{n}-b\frac{\sum x}{n}-a\frac{\sum x^2}{n}$

Where,

$\sum xx=\sum x^2-\frac{(\sum x)^2}{n}$

$\sum xy=\sum xy-\frac{\sum x\sum y}{n}$

$\sum xx^2=\sum x^3-\frac{\sum x\sum x^2}{n}$

$\sum x^2y=\sum x^2y-\frac{\sum x^2\sum y}{n}$

$\sum x^2x^2=\sum x^4-\frac{(\sum x^2)^2}{n}$

Putting the values, we get

$a=14.9786,b=3.1057,c=56.2771$

Putting these in the quadratic equation,

$y=14.9786x^2+3.1057x+56.2771$

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.

$y(10)=14.9786(10)^2+3.1057(10)+56.2771=1585.1941 \approx 1586$