Question

Find the horizontal or oblique asymptote of f(x) = negative 2 x squared plus 3 x plus 6, all over x plus 1.

Answer 1

medal and fan to whoever will tell me the CORRECT answer

Answer 2

@mathmale @Kitten_is_back @winsteria @daniel12900 @you_me @nonopro

Answer 3

Hello. Your equation has some ambiguity to it. You've written
"f(x) = negative 2 x squared plus 3 x plus 6, all over x plus 1."
May I verify this? \[f(x)=\frac{ (-2x)^2+3x+6 }{ x+1 }\]

Answer 4

Make any necessary corrections before we proceed further.

Answer 5

correct

Answer 6

Note that this is a RATIONAL function. How does one determine whether or not a rational function has vertical asymptotes and/or horizontal asymptotes?

Answer 7

i dunno that's why im asking you haha

Answer 8

Please note that neither helpers nor learners are allowed to exchange "correct answers" on OpenStudy. Helpers help without giving away answers. So do not offer "medal and fan" to anyone who gives you the "correct answers."

Is yours an online course? If so, isn't there a discussion of horizontal and vertical asymptotes there?

Answer 9

Once you understand how horiz. and vert. asymptotes are determined, finding them is often a snap. That's the case here.

Horizontal asymptotes are determined by taking the limit as x grows very large of the rational expression you've started with.

Vertical asymptotes are determined by setting the denom. of your rational fn. to 0 and solving for the x-value(s).

Answer 10

Long division is one approach for determining the equations of slant asymptotes.

Answer 11

i have absolutely no clue on what you are talking about

Answer 12

I cannot tell you the answer, but I can help you through the question.
You see, horizontal asymptotes are usually calculated with the limit: \(\lim_{x \rightarrow \pm \infty} f(x)\)
And depending on this limits result we can draw some conclusions, but the scenarios are as follow:
\((I)\) if \(\lim_{x \rightarrow \pm \infty} f(x) = b\) then the asymptote will be horizontal and with equation \(y=b\).
\((II)\) if \(\lim_{x \rightarrow \pm \infty} f(x)= \pm \infty\) then we should study the limit of \(\lim_{x \rightarrow \pm \infty} \frac{ f(x) }{ x }\) which can yield three scenarios:
\((i)\)\(\lim_{x \rightarrow \pm \infty} \frac{ f(x) }{ x } = \pm \infty \) then the asymptote is parallel to the "y" axis.
\((ii) \lim_{x \rightarrow \pm \infty}\frac{ f(x) }{ x } = 0\) then the asymptote is parallel to the x-axis which also implies there is no asymptote (usually falls exactly on the x-axis).
\((iii) \lim_{x \rightarrow \pm \infty} \frac{ f(x) }{ x }= m\) where \(m \in \mathbb{R}\) then we have to study the limit \(\lim_{x \rightarrow \pm \infty} f(x)-mx\) in order to find the oblique asymptote, this at the same time can yield two different outcomes:
\((iii_1) \lim_{x \rightarrow \pm \infty} f(x)-mx = \pm \infty\) this implies that the asymptote has direction of the slope "m"
\((iii_2) \lim_{x \rightarrow \pm \infty }f(x)-mx= b\) then the asymptote will be defined by the line \(y=mx+b\)

Answer 13

so it would be y=2x+3

Answer 14

I need to see how you obtained that possible result.

Answer 15

\[f(x)=\frac{ (-2x)^2+3x+6 }{ x+1 }\]
can be re-written as
\[f(x)=\frac{ 4x^2+3x+6 }{ x+1 }\]

Answer 16

Assuming that you're familiar with "limits at infinity," please determine the limiting value of this f(x) as x increases without bound. You can actually do that by inpection; no calculations necessary.

Answer 17

i just pulled out y=mx + b

Answer 18

That's the equation of a straight line. It has some relevance here, but does not answer the question that you've posted.

Have you heard of and / or used "limits at infinity?"

Answer 19

nope

Answer 20

Looks as tho' you'lll need to do some reading before the solution of this problem will make sense to you. Is yours an online course? Have you any online references that explain "limits at infinity?" or a textbook or workbook?

Answer 21

@KirbyLegs

Answer 22

@scenemunster : If you truly want to understand and remember the math concepts being discussed here, you'll need to get personally involved. Again I urge you to look up "limits at infinity" online, in your textbook (if you have one) or notes (if you have any).

Ask yourself: if you allow x to go to infinity (grow larger without bounds) in the following function, what limiting value does the function have?

\[f(x)=\frac{ 4x^2+3x+6 }{ x+1 }\]

Answer 23

Hint: Divide both numerator and denominator by the highest power of x, which is x^2. type in the resulting function. Then let x approach infinity. What limiting value does the function f reach? Hint: the correct answer is not a number; it's a symbol for "increases without bound."

Answer 24

A first step:
\[\lim_{x \rightarrow \infty} \frac{ 4x^2+3x+6 }{ x+1 } \iff \lim_{x \rightarrow \infty} \frac{ 4x^2 }{ x }\]

Answer 25

Should I be using limits for this @mathmale ?
She does not seem to know about operating limits.

Answer 26

@scenemunster: Kindly tell us the name of the course you're in. That would help us to suggest an approach suitable for your course. @Owlcoffee and I would use the method of limiits, but if you're not studying that, there are other ways in which to find the answer to the question you've posted. You could, for example, graph the function. Or you could divide every term in the given function by the highest power (x^2) and then ask yourself what happens to each of the resulting terms if x grows larger and larger.

Answer 27

@scenemunster: OpenStudy says you're offline. If you wish, continue this discussion by using some of my suggestions when you're back online.