Linear-quadratic graph intersections (ACT). Game on.

Set the expressions equal: $3x-1=x^{2}+x-1$, so $0=x^{2}-2x$. Then $x(x-2)=0$, and $x=0$ or $x=2$. For $x=2$, $y=3(2)-1=5$, so one intersection point is $(2,5)\textsf{.}$

Set $f(x)=g(x)$: $-x+5=x^{2}-3x+5$, so $0=x^{2}-2x$. Then $x(x-2)=0$, and $x=0$ or $x=2$. For $x=2$, $f(2)=3$, so one intersection point is $(2,3)\textsf{.}$

Set the expressions equal: $x-4=-x^{2}+5x-4$, so $0=-x^{2}+4x$. Then $0=-x(x-4)$, and $x=0$ or $x=4$. For $x=4$, $y=4-4=0$, so one intersection point is $(4,0)\textsf{.}$

Set $f(x)=g(x)$: $2x+3=x^{2}-x+3$, so $0=x^{2}-3x$. Then $x(x-3)=0$, and $x=0$ or $x=3$. For $x=0$, $f(0)=3$, so one intersection point is $(0,3)\textsf{.}$

Set the expressions equal: $-3x+2=x^{2}-4x+2$, so $0=x^{2}-x$. Then $x(x-1)=0$, and $x=0$ or $x=1$. For $x=1$, $y=-3(1)+2=-1$, so one intersection point is $(1,-1)\textsf{.}$

Set the expressions equal: $6-2x=x^{2}-5x+6$, so $0=x^{2}-3x$. Then $x(x-3)=0$, and $x=0$ or $x=3$. For $x=3$, $y=6-2(3)=0$, so one intersection point is $(3,0)\textsf{.}$

Set $p(x)=q(x)$: $\dfrac{1}{2}x+1=x^{2}-\dfrac{1}{2}x-1$, so $0=x^{2}-x-2$. Then $(x-2)(x+1)=0$, and $x=2$ or $x=-1$. For $x=-1$, $p(-1)=\dfrac{1}{2}$, so one solution is $\left(-1,\dfrac{1}{2}\right)\textsf{.}$

Set the expressions equal: $4x-3=2x^{2}-x-3$, so $0=2x^{2}-5x$. Then $x(2x-5)=0$, and $x=0$ or $x=\dfrac{5}{2}$. For $x=\dfrac{5}{2}$, $y=4\left(\dfrac{5}{2}\right)-3=7$, so one intersection point is $\left(\dfrac{5}{2},7\right)\textsf{.}$