sketch the graph of f(x)=x^2+2x-8/x^2+6x+8 lim f(x) x=-4

Question

limit of f(x) x extends to -4

Answer 1

me thank yu wanna EVALUATE dat dang thang at x=-4 . . . its pretty much x^2/x^2, so gotta be kloes tu 1 . . . top=x^2+2x-8, at x=-4...16-8-8=0 . . . anser=0

Answer 2

You can see what it looks like by entering this into Google:  plot((x^2+2x-8)/(x^2+6x+8))

Make sure you zoom out using the mouse wheel.  There are two parts to the graph.

Note:

(x^2 + 2x - 8) / (x^2 + 6x + 8)

(x+4)(x-2) / (x+4)(x+2)

(x-2)/(x+2) with a hole knocked out at x = -4

Since you can't have 0 on the bottom of a fraction, x+2 can't be 0, so x can't be -2.

There's a vertical asymptote at x = -2.