If it's a quadratic we know that it must look like
for some numbers a,b and c. To find these numbers we use the coordinates.
(0,-2) lies on the curve so when x=0, y=-2
![y=ax^2+bx+c \Rightarrow -2=a(0^2)+b(0)+c \Rightarrow -2 = c](https://tex.z-dn.net/?f=%20y%3Dax%5E2%2Bbx%2Bc%20%5CRightarrow%20-2%3Da%280%5E2%29%2Bb%280%29%2Bc%20%5CRightarrow%20-2%20%3D%20c%20)
(so that solves one of them)
Next (1,0) lies on the curve so when x=1, y=0
![y=ax^2+bx-2 \Rightarrow 0=a(1^2)+b(1) -2 \Rightarrow a+b-2=0 \Rightarrow a+b=2](https://tex.z-dn.net/?f=%20y%3Dax%5E2%2Bbx-2%20%5CRightarrow%200%3Da%281%5E2%29%2Bb%281%29%20-2%20%5CRightarrow%20a%2Bb-2%3D0%20%5CRightarrow%20a%2Bb%3D2%20)
Finally (3,10) lies on the curve so when x=3, y=10
![y=ax^2+bx-2 \Rightarrow 10 = a(3^2)+b(3)-2 \Rightarrow 9a+3b-2=10 \Rightarrow 9a+3b=12 \Rightarrow 3a+b=4](https://tex.z-dn.net/?f=%20y%3Dax%5E2%2Bbx-2%20%5CRightarrow%2010%20%3D%20a%283%5E2%29%2Bb%283%29-2%20%5CRightarrow%209a%2B3b-2%3D10%20%5CRightarrow%209a%2B3b%3D12%20%5CRightarrow%203a%2Bb%3D4%20)
So then we have to solve the simultaneous equations to get a and b
![a+b=2 \\ 3a+b=4](https://tex.z-dn.net/?f=%20a%2Bb%3D2%20%5C%5C%203a%2Bb%3D4%20%20)
These equations have solutions a=1 and b=1 and so we have a=1, b=1, c=-2.
Therefore the equation of the quadratic is ![y=x^2+x-2](https://tex.z-dn.net/?f=%20y%3Dx%5E2%2Bx-2%20)