Question

Find the horizontal or oblique asymptote of f(x) = 2x^2 + 5x + 6/x + 3

Answer 1

Answer choices:

Answer 2

@Destinymasha @phi @Punkchick @Luigi0210 @Callisto @marylou004 @emily_grace2 Help me out please :)

Answer 3

do long devision

Answer 4

How? lol

Answer 5

@satellite73 @ehuman @wolf1728 Can you help me out please?

Answer 6

No..

Answer 7

That's all that it asks.

Answer 8

Okay, thanks :)

Answer 9

Perhaps @broken_symmetry @superdavesuper @dresdendrew can help :)

Answer 10

Okay, so how would I get the answer after that?

Answer 11

@satellite73 Help?

Answer 12

For the function you quoted, there is one vertical asymptote at x=0 and no others.

Did you mean
\[y = \frac{2x^2 + 5x + 6}{x + 3}\]
instead of
\[y = 2x^2 + 5x + \frac6x + 3\]?

Answer 13

Yes, I did

Answer 14

@broken_symmetry

Answer 15

Have you tried dividing the polynomial?
Try to get something of the form
\[(Ax + B)(x+3) = 2x^2 + 5x+6-C\]

This will mean that
\[y = (Ax+B)+ \frac{C}{x+3}\]
The function Ax+B will be the asymptote you're looking for

Answer 16

Okay, I'm following. Can you show me how to get the answer? :)

Answer 17

Multiply out (Ax+B)(x+3)

Then compare coefficients of x^2, x and 1 with the right hand side

Answer 18

So, it'll be y = -5? I haven't done this stuff for weeks, I'm probably wrong :/

Answer 19

No, what I mean is do
\[(Ax+B)(x+3) = Ax^2+(B+3A)x+3B = 2x^2 + 5x + 6 - C\]
This tells you
A = 2
3A+B = 5
3B = 6-C
The function Ax+B is your answer because if
\[y = (Ax+B)+\frac{C}{x+3}\]
then
\[\lim_{x\to\pm\infty} y \to Ax+B\]

Answer 20

Oh okay, so did I get it right? What's the answer? :)

Answer 21

Can you please just give me the answer? I've been working on this one problem for an hour!

Answer 22

Figure it out?