# Thread: Finding asympotes, turning point and domain!

1. ## Finding asympotes, turning point and domain!

Given the function $f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta}$, state the equations of all the asymptotes and find the turning points, stating the domain.

Please show complete working out. I don't know how to find out the above.

2. domain of function is region for which function is defined, so solve
$x^2 - (\alpha + \beta) + \alpha \cdot \beta = 0$

from there you will go to asymptotes, vertical, horizontal .....

3. Originally Posted by Joker37
Given the function $f(x)=\frac{A}{x^2-(\alpha +\beta)+\alpha\beta}$, state the equations of all the asymptotes and find the turning points, stating the domain.

Please show complete working out. I don't know how to find out the above.
Should the sum of the roots be multiplied by x ?

4. Originally Posted by Archie Meade
Should the sum of the roots be multiplied by x ?
Sorry, yeah you're right. I made a mistake. Below is the correct equation:
$f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta}$,

5. Ok,

notice that

$x^2-(\alpha+\beta)x+\alpha\;\beta=(x-\alpha)(x-\beta)$

This shows you the vertical asymptotes, as the values of x that cause the denominator to be zero
are evident.

The horizontal asymptote is discovered by letting x approach infinity.

6. Originally Posted by Archie Meade
Ok,

notice that

$x^2-(\alpha+\beta)x+\alpha\;\beta=(x-\alpha)(x-\beta)$

This shows you the vertical asymptotes, as the values of x that cause the denominator to be zero
are evident.

The horizontal asymptote is discovered by letting x approach infinity.
Makes sense. What about the domain?

7. Originally Posted by Joker37
Makes sense. What about the domain?
The function is defined for all real x except for the values of x that cause the denominator to be zero.
(The vertical asymptotes).
You see these from the factored version of the denominator.

Then you need the turning points, discovered from $f'(x)=0.$

8. Archie Meade, I'm having trouble finding the turning point. I'm wondering if you could post up a complete step-by-step working out to it. Any more help will be appreciated!

9. Your previous method of partial fractions is useful for calculating the derivative of quotients.

Alternatively, use the Quotient Rule.

However, in this case, it's convenient to use a "negative exponent" instead.

$f(x)=\displaystyle\frac{A}{x^2-(\alpha+\beta)x+\alpha\;\beta}=A\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^{-1}$

Using the Chain Rule

$u=x^2-(\alpha+\beta)x+\alpha\;\beta$

$\displaystyle\frac{d}{dx}Au^{-1}=A\frac{du}{dx}\;\frac{d}{du}u^{-1}$

$=\displaystyle\ A\left[2x-(\alpha+\beta)\right]\left(-\left[u^{-2}\right]\right)=\frac{-A\left[2x-(\alpha+\beta)\right]}{\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^2}$

This is zero when the numerator is zero, if we do not have an $(x-\alpha)$ or $(x-\beta)$ factor there.

(all the work calculating the denominator is often a waste of time, but not always).

Since $A\ne\ 0$

$\displaystyle\ 2x-(\alpha+\beta)=0\Rightarrow\ x=\frac{\alpha+\beta}{2}$

There is a turning point at that x, provided $\alpha\ne\ \beta$,

otherwise, we'd have calculated a turning point at the asymptote, which is not possible.

Then place that x into $f(x)$ to find the vertical co-ordinate of the turning point.

10. Okay, thanks. I'm copying and pasting this on word to print for future reference!

11. Originally Posted by Archie Meade
Since $A\ne\ 0$

$\displaystyle\ 2x-(\alpha+\beta)=0\Rightarrow\ x=\frac{\alpha+\beta}{2}$

There is a turning point at that x, provided $\alpha\ne\ \beta$,

otherwise, we'd have calculated a turning point at the asymptote, which is not possible.
How do you know that A does not equal 0?
And why aren't we solving:

$
-A[2x-(\alpha+\beta)]=0
$
?

Where does the A go?

12. Originally Posted by Joker37
How do you know that A does not equal 0?
And why aren't we solving:

$
-A[2x-(\alpha+\beta)]=0
$
?

Where does the A go?
look at the original function ... what would f(x) equal if A = 0 ?

since A cannot equal 0, divide both sides of the equation by -A ...

$\frac{-A[2x-(\alpha + \beta)]}{-A} = \frac{0}{-A}$

$2x-(\alpha + \beta) = 0$

13. Originally Posted by skeeter
look at the original function ... what would f(x) equal if A = 0 ?

since A cannot equal 0, divide both sides of the equation by -A ...

$\frac{-A[2x-(\alpha + \beta)]}{-A} = \frac{0}{-A}$

$2x-(\alpha + \beta) = 0$
Alright, alright, makes sense. I don't really pretend to be good at all this maths stuff.

Then place that x into $f(x)$ to find the vertical co-ordinate of the turning point.
I'm also not getting the correct corresponding y value when I algebraically substitute the x value of the turning point in f(x). Would anyone be able to supply the rest of the working out to see where I went wrong?

14. $f(\frac{\alpha+\beta}{2}) = A[(\frac{\alpha+\beta}{2})^2 -(\alpha+\beta)+\alpha\beta]^{-1}
=A[\frac{\alpha^2 +\beta^2}{4} - \frac{4(\alpha+\beta)}{4} +\frac{4\alpha\beta}{4}]^{-1}$

$=A[\frac{\alpha^2+\beta^2-4\alpha+4\beta+4\alpha\beta}{4}]^{-1}$
Where have I gone wrong?

15. Originally Posted by Joker37
$f(\frac{\alpha+\beta}{2}) = A[\left(\frac{\alpha+\beta}{2}\right)^2 -(\alpha+\beta)+\alpha\beta]^{-1}
=A[\frac{\alpha^2 +\beta^2}{4} - \frac4{\alpha+\beta}{4} +4{\alpha\beta}{4}]^{-1}$

$=A[\frac{\alpha^2+\beta^2-4\alpha+4\beta+4\alpha\beta}{4}]^{-1}$

Where have I gone wrong?
you dropped the x again!

$\displaystyle\ f\left(\frac{\alpha+\beta}{2}\right)=A\left[\left(\frac{\alpha+\beta}{2}\right)^2-(\alpha+\beta)\left[\frac{\alpha+\beta}{2}\right]+\alpha\;\beta\right]^{-1}$

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