Question 690021
w-1 = 0


w = 1


So the vertical asymptote is w = 1


Plug in w = 0 to get (w^2+2w+4)/(w-1) = ((0)^2+2(0)+4)/(0-1) = -4.


So the y-intercept is (0,-4)


Since the discriminant of w^2+2w+4 is negative, there are no real solutions, which means that there are no x-intercepts.


Use polynomial long division to get the quotient w+3. So the oblique asymptote is y = x+3


So what we have is this



{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,0,(x^2+2x+4)/(x-1),x+3),
 blue(line(1,-20,1,20))),
 blue(line(-20,-17,20,23)))

)}}}


Note: The blue vertical line should be a dashed line since it's the vertical asymptote. The blue diagonal line should be a dashed line since it's the oblique asymptote.