To find the x-coordinates of A and B, find the zeroes of the equation (set y=0 and solve for x).
[tex]y=3x+2-2x^2[/tex]
If y=0 then:
[tex]0=3x+2-2x^2[/tex]
Writing this quadratic equation in standard form, we get:
[tex]2x^2-3x-2=0[/tex]
Use the quadratic formula to find the solutions for x:
[tex]\begin{gathered} \Rightarrow x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4(2)(-2)}}{2(2)} \\ =\frac{3\pm\sqrt[]{9+16}}{4} \\ =\frac{3\pm\sqrt[]{25}}{4} \\ =\frac{3\pm5}{4} \\ \Rightarrow x_1=\frac{3+5}{4}=\frac{8}{4}=2 \\ \Rightarrow x_2=\frac{3-5}{4}=\frac{-2}{4}=-\frac{1}{2} \end{gathered}[/tex]
Then, the x-coordinate of A is -1/2, and the x-coordinate of B is 2. Both the y-coordinate of A and B are 0.
On the other hand, to find the y-coordinate of C, which is the point where the graph crosses the Y-axis, replace x=0:
[tex]\begin{gathered} y=3(0)+2-2(0)^2 \\ =2 \end{gathered}[/tex]
Therefore, the coordinates of A, B and C are:
[tex]\begin{gathered} A(-\frac{1}{2},0) \\ B(2,0) \\ C(0,2) \end{gathered}[/tex]
