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#### confusedatmath

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- Jan 2, 2014

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well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....

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- Feb 15, 2012

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Let's see what happens at $x = 1,000,000$, when $a = b = -1,c = 2$.

Then $f(x) = \dfrac{-1}{(999,999)^2} + 2$

$= \dfrac{-1}{999,998,000,001} + 2$

$\sim 1.999999999998999998$

At $x = -1,000,000$, we get:

$f(x) = \dfrac{-1}{(-1,000,001)^2} + 2$

$= \dfrac{-1}{1,000,002,000,001} + 2$

$\sim 1.999999999999000002$

Both of these numbers are really close to 2, right?

In general, we see that:

$\displaystyle \lim_{x \to \infty} \left(\frac{a}{(x + b)^2} + c\right)$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x + b)^2} + \lim_{x \to \infty} c$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x+b)^2} + c$

$\displaystyle = (a)\left(\lim_{x \to \infty}\frac{1}{x+b}\right)^2 + c$

$\displaystyle = (a)(0)^2 + c = 0 + c = c$

Similar reasoning holds to show that:

$\displaystyle \lim_{x \to -\infty} \left(\frac{a}{(x + b)^2} + c\right) = c$

as well.

(If you haven't been formally introduced to limits yet, all you need to know for this is the following (which hold under "suitably nice conditions" which are the case here):

1) the limit of a sum is the sum of the limits of each term in the sum

2) the limit of a product is the product of the limits of each factor in the product

3) if M(x) is a function that gets "infinitely big" as x does, then:

$$\lim_{x \to \infty} \frac{1}{M(x)} = 0$$

-by "get infinitely big as x does" I mean that for ANY positive integer $K$, there is always some positive integer $N$, so that if we have $x > N$, then $|M(x)| > K$ (typically, the integer $N$ will depend on $K$, bigger $K$'s usually need bigger $N$'s).).

The short answer, if [tex]\displaystyle \begin{align*} y = c \end{align*}[/tex] then you end up with [tex]\displaystyle \begin{align*} 0 = \frac{a}{(x + b)^2} \end{align*}[/tex]. Is that possible to solve for x?I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?

View attachment 1838

The long answer, consider the inverse relation. You should find that c becomes a vertical asymptote.

- Jan 29, 2012

- 1,151

The question was "what happens when x is large?" "2" is NOT large!

well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....

If x= 1000000, then 1/(1+ 1000000)+3= 3.000009999900000999990000099999 and for