Construct a function given two asymptotes

[Rectifier]

The problem
Giva an example of a function $f(x)$ that has one vertical asymptote at $x = -1$ and
and another asymptote that is $y=8x+7$.

Translated from Swedish.

The attempt
I know that I should use the hyperbola here but I am not sure how to adapt the hyporbola to the tilting asymptote. Can anyone help please?

Hyperbola:
$(\frac{x}{a})^2-(\frac{y}{b})^2 = 1$
or
$(\frac{y}{b})^2-(\frac{x}{a})^2 = 1$
(I gess that doesn't matter which one I choose since both can satisfy our demands. )

My first thought was to move the whole function f 7 units of length up ,thus the function we are looking fore (lets call it g) is f(x) +7. I now have to adjust the formula above to the tilt of 8.

$(\frac{x}{a})^2-(\frac{y}{b})^2 = 1 \\ (\frac{y}{b})^2 =(\frac{x}{a})^2- 1 \\ (\frac{y}{b}) = \pm \sqrt{(\frac{x}{a})^2- 1} \\$
1 goes away for when x-> $\infty$

$(\frac{y}{b}) = \pm \sqrt{(\frac{x}{a})^2- 1} \\ \frac{y}{b} = \pm \sqrt{(\frac{x}{a})^2}$

There are 2 asymptotes here

$\frac{y}{b} = \sqrt{ (\frac{x}{a})^2 } \\ y = \frac{xb}{a}$

and

$\frac{y}{b} = - \sqrt{ (\frac{x}{a})^2 } \\ y = - \frac{xb}{a}$

the tilt (k) is thus

$k=\frac{b}{a} \\ 8=\frac{b}{a}$

or
the tilt (k) is thus

$k = \frac{-b}{a} \\ 8 = \frac{-b}{a}$

We can pick a=1 and adjust b accordingly.
$8 = -\frac{b}{1} \\ -8 = b$A hyporbola where a = 1 and b=-8 does satisfy (hopefully :) ) our requierments.
$(x)^2-( \frac{y}{-8} )^2 = 1$

this one is not a function though so I rearranged the the formula (removed one half of the range not sure if there is a proper word for it) and got the following function
$$ f(x) = 8 \sqrt{x^2-1} $$

therefore

$$ g(x) = 8 \sqrt{x^2-1} + 7 $$
but fore some reason that was wrong...

Can someone please help?

 

[ehild]

Does your function $g(x) = 8 \sqrt{x^2-1} + 7$ have a vertical asymptote at x = -1?

Do not stick to the hyperbola.
What can be a very simple function g(x) which has a vertical asymptote at x=-1? So as g(x) goes to +or - infinity when x-->-1?
You get an asymptote y=8x+7 by using a factor that tends to 1 when x goes to infinity. Use g(x) in this factor.

 

[SammyS]

The problem
Giva an example of a function $f(x)$ that has one vertical asymptote at $x = -1$ and
and another asymptote that is $y=8x+7$.

Translated from Swedish.

The attempt
I know that I should use the hyperbola here but I am not sure how to adapt the hyperbola to the tilting asymptote. Can anyone help please?

Hyperbola:
$(\frac{x}{a})^2-(\frac{y}{b})^2 = 1$
or
$(\frac{y}{b})^2-(\frac{x}{a})^2 = 1$
(I gess that doesn't matter which one I choose since both can satisfy our demands. )
...
...
therefore,$$ g(x) = 8 \sqrt{x^2-1} + 7 $$
but fore some reason that was wrong...

Can someone please help?

For get the hyperbola. It doesn't have a vertical asymptote.

Try a rational function. One which has the desired vertical asymptote ans also has a slant asymptote.

 

[Rectifier]

Thank you for your help!

For get the hyperbola. It doesn't have a vertical asymptote.

Try a rational function. One which has the desired vertical asymptote ans also has a slant asymptote.

Forgot that I had that other asymptote too...

$f(x)= \frac{1}{x+1}$

has a vertical asymptote but I am not sure how to get that slant asymptote though...

 

[SammyS]

Thank you for your help!
Forgot that I had that other asymptote too...

$f(x)= \frac{1}{x+1}$

has a vertical asymptote but I am not sure how to get that slant asymptote though...

What is the behavior of $\displaystyle \ f(x)= \frac{1}{x+1}\$ as $\displaystyle \ x \to \pm \infty \ ?$

 

[Rectifier]

What is the behavior of $\displaystyle \ f(x)= \frac{1}{x+1}\$ as $\displaystyle \ x \to \pm \infty \ ?$

f(x) -> 0

 

[SammyS]

f(x) -> 0

That's correct.

What happens if you add the function $\ h(x) =8x+7\$ to $\ f(x)\ ?$

 

[Rectifier]

That's correct.

What happens if you add the function $\ h(x) =8x+7\$ to $\ f(x)\ ?$

Add like + or should I multiply it?

 

[HallsofIvy]

Try thinking about this! If you were to multiply the two functions what would you get as x goes to infinity? What if you added then?

 

[Rectifier]

Try thinking about this! If you were to multiply the two functions what would you get as x goes to infinity? What if you added then?

Thank you for the reply.

I have seriously no idea of what would hapen (I guess that there is no easy answer since the behaviour of the product is so unpredictable - at least for me) :,( I have tried to simplify it with same functions like f(x)=x and g(x)=x f(x)g(x)=x^2 but there is no good easy answer. Its easier when you add stuff since then you just add the values from each function together for all x-es. The easiest case is when you add a constant function since you basically move the graph up or down.

 

[SammyS]

Thank you for the reply.

I have seriously no idea of what would hapen (I guess that there is no easy answer since the behaviour of the product is so unpredictable - at least for me) :,( I have tried to simplify it with same functions like f(x)=x and g(x)=x f(x)g(x)=x^2 but there is no good easy answer. Its easier when you add stuff since then you just add the values from each function together for all x-es. The easiest case is when you add a constant function since you basically move the graph up or down.

Try it each way .

For addition: If f(x) → 0 for large x, then what is the effect at large x, if you add f(x) to some other function?

For multiplication: If f(x) → 0 for large x, and you multiply it by a function that goes to ±∞, then you need to investigate further, which isn't too difficult. Otherwise, what do you suppose happens to the product (multiplication) for large x ?