Question

nothing

Answer 1

Your answer is correct.

Answer 2

Do you know how to work these problems?

Answer 3

Factor the numerator and denominator and see if you can cancel any terms.

Here, you can't, but you can divide the numerator by the denominator to find the oblique asymptote. What do you get?

Answer 4

You also know that the graph has a hole or vertical asymptote wherever the denominator equals 0.

Answer 5

That rules out the choice you made, unfortunately...

Answer 6

Can you do the long division of \[x^2+3x+5\]------------\[x+2\]?

Answer 7

What's the first term of the quotient going to be?

Answer 8

very good. so there's going to be an invisible line with the equation y = x+1 that the graph "hugs"...

Answer 9

that's your oblique asymptote or slant asymptote

Answer 10

So, sketch that line on your graph paper, so you know where the graph cannot go (you never cross an asymptote).

Answer 11

Also sketch the vertical asymptote at x = -2

Answer 12

Yes, I believe D is correct, because the other graph with similar slant asymptote has the wrong vertical asymptote.