# Thread: Asymptotes - Help f(x)=x²-5x+7/(x-2)

1. ## Asymptotes - Help f(x)=x²-5x+7/(x-2)

f(x)=x²-5x+7/(x-2)

It asks me to give the vertical, horizontal and slant asymptotes if there are any.

This is what I have so far:

vertical asymptote: x cannot equal 2

I am now trying to find the slant asymptote by synthetic (long division) because x²-5x+7 does not factor out.

I am really stuck so may someone show me how to do this problem? I still need to find the slant and horizontal asymptotes.

2. Originally Posted by agentlopez
f(x)=x²-5x+7/(x-2)

It asks me to give the vertical, horizontal and slant asymptotes if there are any.

This is what I have so far:

vertical asymptote: x cannot equal 2

I am now trying to find the slant asymptote by synthetic (long division) because x²-5x+7 does not factor out.

I am really stuck so may someone show me how to do this problem? I still need to find the slant and horizontal asymptotes.
The vertical asymptote is not "x cannot equal 2" it is the line x = 2.

$\frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

So the slant asymptote is $y = x - 3$.

For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.

-Dan

3. Originally Posted by topsquark
The vertical asymptote is not "x cannot equal 2" it is the line x = 2.

$\frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

So the slant asymptote is $y = x - 3$.

For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.

-Dan
Can you show me how you got $y = x - 3$ and what would I get as the horizontal asymptote? I know you used synthetic division, but I need help on how to get to $y = x - 3$ and what would be my horizontal asymptote?

4. You don't need synthetic division:

$\frac{{x^2 - 5x + 7}}
{{x - 2}} = \frac{{x^2 - 5x + 6 + 1}}
{{x - 2}} = \frac{{(x - 2)(x - 3) + 1}}
{{x - 2}} = x - 3 + \frac{1}
{{x - 2}}.$

For horizontal asymptotes, evaluate

$\lim_{x \to \infty } \frac{{x^2 - 5x + 7}}
{{x - 2}}$
& $\lim_{x \to -\infty } \frac{{x^2 - 5x + 7}}
{{x - 2}}.$

5. Originally Posted by Krizalid
You don't need synthetic division:

$\frac{{x^2 - 5x + 7}}
{{x - 2}} = \frac{{x^2 - 5x + 6 + 1}}
{{x - 2}} = \frac{{(x - 2)(x - 3) + 1}}
{{x - 2}} = x - 3 + \frac{1}
{{x - 2}}.$

For horizontal asymptotes, evaluate

$\lim_{x \to \infty } \frac{{x^2 - 5x + 7}}
{{x - 2}}$
& $\lim_{x \to -\infty } \frac{{x^2 - 5x + 7}}
{{x - 2}}.$

Now one has ever showed me that you can just split a number and do that. Something new I learned^^. I still do not understand the horizontal asymptote that you are explaining to me. I need someone to be more detailed with this.

Originally Posted by topsquark
$\frac{x^2 - 5x + 7}{x - 2} = x - 3 + \frac{1}{x - 2}$

So the slant asymptote is $y = x - 3$.

For horizontal asymptotes we find the limit of the function as x goes to plus or minus infinity. You can see for yourself that this function blows up/down at positive/negative infinity. Or you can simply use the fact that a function with a slant asymptote will never have a horizontal asymptote.
--

Now, if you evaluate $x - 3 + \frac{1}{x - 2}$ as $x\to\infty$ or $x\to-\infty,$ does that make sense?

7. Originally Posted by Krizalid

--

Now, if you evaluate $x - 3 + \frac{1}{x - 2}$ as $x\to\infty$ or $x\to-\infty,$ does that make sense?

^^Not yet^^

Maybe I was not specific enough when I asked to be more descriptive. I am a beginner at this stuff and I realise that the best way to learn is to figure it out for yourself, but I am really stuck and need this explained 'THOROUGHLY'. Nothing is 'clicking' for me even when you basically just reworded it above. I need this explained as if this is something completely new to me. I really need this so please help me understand.

What your telling me is all slant asymptotes do not have a horizontal asymptote? I thought you needed the horizontal and vertical asymptotes for when there is a slant asymptote.

8. Think of it this way...

As $x \rightarrow \infty$, $f(x) \rightarrow x-3$

This is because as x goes to infinity, $\frac{1}{x-2}$ becomes smaller and smaller, eventually becoming practically non-existent. But the $x-3$ part keeps rising steadily as $x$ does, so $f(x)$ slowly converges upon $y=x-3$.

9. You don't need synthetic division:

How did you know to split the 7 into 6+1?

10. When I saw Dan's algebra, I was wondering how I could split the original ratio into two sums.

Then clearly we need to find something on the top which yields $x-2.$ I thought in $(x-2)(x-3)$ 'cause the first two terms are exactly $x^2$ & $-5x,$ so from here, we're done, 'cause the third term is $6.$ Since on the numerator we have $x^2-5x+7$ the conclusion follows quickly.

--

This does not require much effort, it's just a way to avoid the looong division

11. Originally Posted by DivideBy0
Think of it this way...

As $x \rightarrow \infty$, $f(x) \rightarrow x-3$

This is because as x goes to infinity, $\frac{1}{x-2}$ becomes smaller and smaller, eventually becoming practically non-existent. But the $x-3$ part keeps rising steadily as $x$ does, so $f(x)$ slowly converges upon $y=x-3$.
Does that mean the horizontal asymptote is 2?

x=2
y=2

f(x)=x-3

Is that right?

12. Originally Posted by agentlopez
Does that mean the horizontal asymptote is 2?

x=2
y=2

f(x)=x-3

Is that right?
Yes, there is a horizontal asymptote at x = 2. Try graphing the function.