we need to write the equation of the graph
it is a parable then the general form is
[tex]y=(x+a)^2+b[/tex]where a move the parable horizontally from the origin (a=negative move to right and a=positive move to left)
and b move the parable vertically from the origin (b=negative move to down and b=positive move to up)
this parable was moving from the origin to the right 2 units and any vertically
then a is -2 and b 0
[tex]y=(x-2)^2[/tex]now we have the system of equations
[tex]\begin{gathered} y=3x-2 \\ y=(x-2)^2 \end{gathered}[/tex]we can replace the y of the first equation on the second and give us
[tex]3x-2=(x-2)^2[/tex]simplify
[tex]3x-2=x^2-4x+4[/tex]we need to solve x but we have terms sith x and x^2 then we can equal to 0 to factor
[tex]\begin{gathered} 3x-2-x^2+4x-4=0 \\ -x^2+7x-6=0 \end{gathered}[/tex]multiply on both sides to remove the negative sign on x^2
[tex]x^2-7x+6=0[/tex]now we use the quadratic formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a is 1, b is -7 and c is 6
[tex]\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(1)(6)}}{2(1)} \\ \\ x=\frac{7\pm\sqrt[]{49-24}}{2} \\ \\ x=\frac{7\pm\sqrt[]{25}}{2} \\ \\ x=\frac{7\pm5}{2} \end{gathered}[/tex]we have two solutions for x
[tex]\begin{gathered} x_1=\frac{7+5}{2}=6 \\ \\ x_2=\frac{7-5}{2}=1 \end{gathered}[/tex]now we replace the values of x on the first equation to find the corresponding values of y
[tex]y=3x-2[/tex]x=6
[tex]\begin{gathered} y=3(6)-2 \\ y=16 \end{gathered}[/tex]x=1
[tex]\begin{gathered} y=3(1)-2 \\ y=1 \end{gathered}[/tex]Then we have to pairs of solutions
[tex]\begin{gathered} (6,16) \\ (1,1) \end{gathered}[/tex]where green line is y=3x-2
and red points are the solutions (1,1)and(6,16)
