The euqaion of a quadratic function is given by:
[tex]y=ax^2+bx+c[/tex]we need to find the values of the constants a, b and c. To do this we use the points given and plug the values of x and y in the equation above.
For the point (-2,0) we have that x=-2 and y=0; then we have:
[tex]\begin{gathered} 0=(-2)^2a+(-2)b+c \\ 4a-2b+c=0 \end{gathered}[/tex]For the point (1,0) x=1 and y=0 then we have the equation:
[tex]\begin{gathered} 0=(1)^2a+(1)b+c \\ a+b+c=0 \end{gathered}[/tex]Finally for the point (0,2) x=0 and y=2 then we have the equation:
[tex]\begin{gathered} 2=(0)^2a+(0)b+c \\ c=2 \end{gathered}[/tex]Hence we have the system of equations:
[tex]\begin{gathered} 4a-2b+c=0 \\ a+b+c=0 \\ c=2 \end{gathered}[/tex]Now to solve this sytem we plug the value of c given by the third equation into the first and second equations to get:
[tex]\begin{gathered} 4a-2b=-2 \\ a+b=-2 \end{gathered}[/tex]To solve this new system we multiply the second equation by 2:
[tex]\begin{gathered} 4a-2b=-2 \\ 2a+2b=-4 \end{gathered}[/tex]we add the equations:
[tex]\begin{gathered} 4a-2b+2a+2b=-2-4 \\ 6a=-6 \end{gathered}[/tex]Now we solve for a:
[tex]\begin{gathered} 6a=-6 \\ a=-\frac{6}{6} \\ a=-1 \end{gathered}[/tex]hence a=-1.
Finally to get b we plug the values of a and c in the first of our originals equations:
[tex]\begin{gathered} 4(-1)-2b+2=0 \\ -4-2b+2=0 \\ -2b-2=0 \end{gathered}[/tex]and solve for b:
[tex]\begin{gathered} -2b=2 \\ b=\frac{2}{-2} \\ b=-1 \end{gathered}[/tex]Hence b=-1.
Now that we have the values of the constant we have that the equation is:
[tex]y=-x^2-x+2[/tex]