find the second degree equation that passes through the points: (0,1) (2,2) (1,0)
first of all we have to recognise the general form of the equation, similar to the equation of a line but we just have to add one more term due to the fact that it is a second degree equation: y=ax2+bx+c. We can see that we have 3 unknowns and 3 points as well, so by replacing the couples of points ("x","y") in the general equation we get the 3 unknowns a,b and c.The value of c is the intersection between the curve and the vertical axis when x=0. When the value of a is positive, the curve will have an upward concavity, if negative it will have a downward concavity. So the process is:the first equation gives us 1=0+0+c, so c=1; the second and third equations are: 2=4a+2b+c, 0=a+b+c. From the last one we get a=-b-c=-b-1 and we replace it in the second one, from where we get b: 4(-b-1)+2b=1 and b=-5/2 the a=3/2.
