Question

Let k be a constant, and consider the function
f(x) = 4x^2 / kx2 + 2x + 1
:
(a) Find all values of k such that f has no vertical asymptotes.

(b) Find all values of k such that f has two vertical asymptotes.
(c) Find the horizontal asymptotees of f, in terms of k. Does f have horizontal asymptotes
for all values of k?

Answer 1

help?

Answer 2

for a is the answer all real numbers except for 0? is k is any number then it cant be factored out thus no 0 in the denominator
So when k = o there is only one vertical asymptote

Answer 3

actually when k = 1 and o it only has one vertical asymptote

Answer 4

0*

Answer 5

\[f(x)=\frac{4x^2}{kx^2+2x+1}\]
\(f(x)\) will have no vertical asymptotes when the denominator is never 0.

You're right about the 1. You can apply the quadratic formula to see if there are any other non-intuitive answers:
\[kx^2+2x+1=0\\
x=\frac{-2\pm\sqrt{4-4k}}{2k}\]
For non-zero k, of course. So what values of \(k\) make this 0?
\[-2\pm\sqrt{4-4k}=0~~\Rightarrow~~k=1\]

Answer 6

Wait, are you looking for one asymptote, or none?

Answer 7

question a asked for none.
I know that there are asymptotes when k= 0 and 1

Answer 8

so I was thinking all real numbers except for 1 and 0

Answer 9

Yes

Answer 10

\[(-\infty,0]\cup [1,\infty)\]
right?

Answer 11

Close. You're not including 1 and 0, but you are including all real numbers besides them. So the interval would be
\[(-\infty,0)\cup(0,1)\cup(1,\infty)\]

Answer 12

I'm stuck on b and c I have an idea but not really

Answer 13

oh okay

Answer 14

\(f(x)\) will have two vertical asymptotes if the bottom could be factored into two distinct factors. First impression is that there are no (real) values of \(k\) that would work.

Answer 15

Actually, I've got one. If \(k=-3\), you have
\[-3x^2+2x+1=(3x+1)(1-x)\]
Checking to see if there are more...

Answer 16

thank you

Answer 17

this looks more complicated than calc 1 lol

Answer 18

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
It looks like there might be plenty more...

\(kx^2+2x+1>0\) for \(k>0\), right? The vertex of this parabola is \(\left(-\dfrac{1}{k},1-\dfrac{1}{k}\right)\). Since \(k>0\), you have a parabola opening upward. If the y-coordinate of the vertex falls under the x-axis (meaning \(1-\dfrac{1}{k}<0\)), then the parabola crosses the x-axis twice. So, this occurs when
\[1-\frac{1}{k}<0~~\Rightarrow~~0<k<1\]Make sense?

For negative \(k\), the vertex is \(\left(\dfrac{1}{k},1+\dfrac{1}{k}\right)\), and it opens downward. So inversely, the parabola would cross the x-axis twice if \(1+\dfrac{1}{k}>0\), or \(k<-1\).
\(\color{blue}{\text{End of Quote}}\)

Answer 19

I sort of understand though

Answer 20

Yeah, I agree, this seems like a bit much for calc I, but it's good practice for reasoning skills. And besides, no calculus knowledge has been used; you should have learned most if not all in pre-calc.

Answer 21

Anyway, it looks like the answer for part (b) is \((-\infty,-1)\cup(0,1)\).

Answer 22

so what about -3?

Answer 23

-3 < -1. It falls in the interval, and it works. It just happens to be the only one (as far as I can tell) that gives you an easily factor-able quadratic.

Answer 24

oh yeah

Answer 25

Now for the last one (which is much simpler, imo), you have to use some calc knowledge.

Horizontal asymptotes occur if the limits at ±∞ of the function are constant.

For example, the function \(g(x)=0\) has a horizontal asymptote of \(y=0\), since the limit as x approaches either end is 0. Sound familiar?

For \(f(x)\), you have to determine the values of \(k\) that give you a constant when you take the limits at infinity.

Answer 26

\[\lim_{x\to\pm\infty}\frac{4x^2}{kx^2+2x+1}=\frac{4}{k}\]
which is a constant for what values of \(k\)?

Answer 27

all real numbers except for 0?

Answer 28

am I right?

Answer 29

I think there may be an easier approach. :) For b, you can again use the quadratic formula.

Assuming \( k \neq 0\) (because it can easily be shown that in that case there is only one VA), if \(\large x=\dfrac{-2\pm\sqrt{4-4k}}{2k}\) gives two distinct, real values for x, then there will be 2 vertical asymptotes.

That will happen if \(\large 4-4k>0\) which means if \(\large 1>k\), and since we already excluded k=0, that means that there are 2 VA if and only if:

\(k<1\) and \(k \neq 0\). Notice that this DOES include all the values of \(k \in (-1,0)\), which you excluded above. I think it's also a simpler approach, just reducing the question to looking at the discriminant.

As for part (c), I don't think calculus is needed here, if you learned the rules for horizontal asymptotes of rational functions (which I know I teach in Algebra 2):

If degree of num'r< degree of den'r, then there is a horizontal asymptote at y=0 (the x axis).

If degree of num'r = degree of den'r, then there is a HA at y=p/q, where p and q are the leading coefficients of the num'r and den'r, respectively.

If degree of num'r > degree of den'r, there are no HA (but will have an oblique asymptote).

So here, you can never have the y=0 HA since deg den'r is either 1 (if k=0) or 2 (if k is not =0).

If \(k \neq 0\) then you are in the "case 2" situation, and the HA is \(y=\dfrac{4}{k}\).
And if k=0, then no HA.

Answer 30

I just looked closer at the discussion of (a) above, and I don't follow the reasoning. You found that if k=1 then the discriminant = 0, but that just tells you where there is a repeated, real root (so 1 VA). So the k=1 and k=0 cases both give exactly ONE VA, but the rest of the possibilities for k includes the 2 VA cases AND the NO VA cases.

There will be NO VA when \(\large x=\dfrac{-2\pm\sqrt{4-4k}}{2k}\) has NO real solutions. That is true when \(\large 4-4k<0\), which is when \(\large 1<k\) (again, ignoring the k=0 case, since we know that gives 1 VA).

So altogether we have:

\(\large k \neq 0,k>1\) means NO VA.
\(\large k \neq 0,k<1\) means TWO VA.
\(\large k=0~or~k=1\) means ONE VA.

And that covers all possibilities for k (actually more than the question asked, lol)