## Calculus: Early Transcendentals (2nd Edition)

$y=x^{2}+2$ $y=x+4$ $x^{2}+2=x+4$ $x^{2}-x-2=0$ $(x-2)(x+1)=0$ $x=2,-1$ $y=2+4=6$ $y=-1+4= 3$ (2,6) and (-1,3)