Answer:
[tex]\large\boxed{x-intercepts:-2\ and\ 4}[/tex]
Step-by-step explanation:
[tex]m(x)=\dfrac{x^2-2x-8}{x^2-x-2}\\\\\text{At the beginning we de}\text{fine the domain:}\\\\D:x^2-x-2\neq0\\\\x^2-2x+x-2\neq0\\x(x-2)+1(x-2)\neq0\\(x-2)(x+1)\neq0\iff x-2\neq0\ \wedge\ x+1\neq0\\\boxed{x\neq2\ \wedge\ x\neq-1}\\================================\\\text{The x-intercept is for}\ m(x)=0.\ \text{Therefore we have the equation:}\\\\\dfrac{x^2-2x-8}{x^2-x-2}=0\iff x^2-2x-8=0\\\\x^2+2x-4x-8=0\\x(x+2)-4(x+2)=0\\(x+2)(x-4)=0\iff x+2=0\ \vee\ x-4=0\\\\x=-2\in D\ \vee\ x=4\in D[/tex]
