First Principles AI

Solution

To find intersection points of $y=f(x)$ and $y=g(x)$ , solve:

f(x)=g(x)

and then substitute back to get $y$ .

\begin{aligned} 2x+5&=-x+5\\ 3x&=0\\ x&=0 \end{aligned}

y=f(0)=2(0)+5=5

\boxed{(0,5)}

\begin{aligned} 3x-2&=10-x\\ 4x&=12\\ x&=3 \end{aligned}

y=f(3)=3(3)-2=7

\boxed{(3,7)}

\begin{aligned} 2x-4&=3x-1\\ -x&=3\\ x&=-3 \end{aligned}

y=f(-3)=2(-3)-4=-10

\boxed{(-3,-10)}

\begin{aligned} -3x-4&=\frac{1}{2}x+3\\ -6x-8&=x+6\\ -7x&=14\\ x&=-2 \end{aligned}

y=f(-2)=-3(-2)-4=2

\boxed{(-2,2)}

\begin{aligned} x-1&=x^2-4x+3\\ 0&=x^2-5x+4\\ 0&=(x-1)(x-4) \end{aligned}

So $x=1$ or $x=4$ .

\begin{aligned} x=1&\Rightarrow y=f(1)=0\\ x=4&\Rightarrow y=f(4)=3 \end{aligned}

\boxed{(1,0)\;\text{and}\;(4,3)}

\begin{aligned} 3x+4&=x^2+2x-8\\ 0&=x^2-x-12\\ 0&=(x-4)(x+3) \end{aligned}

So $x=4$ or $x=-3$ .

\begin{aligned} x=4&\Rightarrow y=f(4)=16\\ x=-3&\Rightarrow y=f(-3)=-5 \end{aligned}

\boxed{(4,16)\;\text{and}\;(-3,-5)}

\begin{aligned} -2x-1&=x^2-4x\\ 0&=x^2-2x+1\\ 0&=(x-1)^2 \end{aligned}

So $x=1$ .

y=f(1)=-2(1)-1=-3

\boxed{(1,-3)}

\begin{aligned} -x^2-3x+2&=x+6\\ 0&=-x^2-4x-4\\ 0&=x^2+4x+4=(x+2)^2 \end{aligned}

So $x=-2$ .

y=g(-2)=-2+6=4

\boxed{(-2,4)}