Given:
[tex]y=-x^2+x+3\text{ and y=-}\frac{1}{2}x+3[/tex]
To find the points of intersection, we have:
[tex]-x^2+x_{}+3=-\frac{1}{2}x+3[/tex]
Solve for x:
[tex]\begin{gathered} -x^2+x+\frac{1}{2}x=3-3 \\ \\ -x^2+\frac{3}{2}x=0 \\ \\ -x(x-\frac{3}{2})=0 \\ x\text{ =0 or }1.5 \end{gathered}[/tex]
Now input 0 for x in equation 1 and 1.5 for x in equation 2
[tex]\begin{gathered} y=0^2^{}+0+3 \\ y\text{ = 3} \end{gathered}[/tex][tex]\begin{gathered} y\text{ = -}\frac{1}{2}(1.5)+3 \\ \text{ y = -}\frac{3}{4}+3 \\ y\text{ = }\frac{9}{4}=2.25 \end{gathered}[/tex]
Thus, the points of intersection for both lines are:
[tex](1.5,\text{ 2.25) and }(0,\text{ 3)}[/tex]
ANSWER:
(1.5, 2.25) and (0, 3)
