- #1
Rectifier
Gold Member
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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 somone please help?
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 somone please help?
Last edited: