Question 208402
You listed this problem under "Quadratic Equation" so I assume the problem is to find a quadratic equation for the given points.<br>

The basic form of a quadratic equation is
{{{y = ax^2 + bx + c}}}
One way to find the equation is to substitute the coordinates of these points into the basic form. If we take the point (1, 1):
{{{(1) = a(1)^2 + b(1) + c}}}
1 = a + b + c
If we use the point (2, 3):
{{{(3) = a(2)^2 + b(2) + c}}}
3 = 4a + 2b + c
If we use the point (3, 6):
{{{(6) = a(3)^2 + b(3) + c}}}
6 = 9a + 3b + c
At this point we have a system of three equations with three variables (a, b and c). We should be able to solve this for a, b and c. The system is:
1 = a + b + c
3 = 4a + 2b + c
6 = 9a + 3b + c
There are a variety of ways to solve this system: Substitution, Linear Combination (aka Elimination), Cramer's Rule (determinants), and a variety of matrix-based methods. I will use Linear Combination. To start I will subtract the first equation from each of the other two equations. The result of these two subtractions:
2 = 3a + b
5 = 8a + 2b
Now I have a system of two equations of two variables (since the c's have been eliminated).
Now I will subtract twice the first equation from the second. The result of this subtraction:
1 = 2a
Dividing by two we get:
1/2 = a
Now that we have "a" we can substitute for "a" and find "b" and "c". Substituting for a in the second equation of the second system:
5 = 8(1/2) + 2b
5 = 4 + 2b
1 = 2b
1/2 = b
Now we can go back to the first system to find "c". Substituting in "a" and "b" into the second equation of the first system:
3 = 4(1/2) + 2(1/2) + c
3 = 2 + 1 + c
3 = 3 + c
0 = c
Now that we have "a", "b" and "c" we can write our quadratic equation by substituting these values into the basic form:
{{{y = ax^2 + bx + c}}}
Substituting:
{{{y = (1/2)x^2 + (1/2)x + 0}}}
or 
{{{y = (1/2)x^2 + (1/2)x}}}
Not only does this equation work for the three points we used, it also works for the fourth point, (4,10).