[SOLVED] Graph question

• Feb 24th 2009, 10:03 AM
brentwoodbc
[SOLVED] Graph question
whats the difference between say...

$\displaystyle 5/x$ or $\displaystyle 1/x$ what translation does it make to the graph? It was on a test I wrote but I was never really taught it. Is it like a reciprocal multiplied by 5?
• Feb 24th 2009, 10:14 AM
Reckoner
Quote:

Originally Posted by brentwoodbc
whats the difference between say...

$\displaystyle 5/x$ or $\displaystyle 1/x$ what translation does it make to the graph? It was on a test I wrote but I was never really taught it. Is it like a reciprocal multiplied by 5?

If $\displaystyle y=\frac1x,$ then $\displaystyle \frac5x$ is just $\displaystyle 5y\text.$ So the graph is "stretched" vertically, by a factor of 5.

I'll give you a graph in a second. Just let me capture it and upload.
• Feb 24th 2009, 10:19 AM
brentwoodbc
thanks(Yes)
• Feb 24th 2009, 10:26 AM
brentwoodbc
im trying to think of how you would find the invariant points.do you just go back to what the original equation was or?

like with 5/x what are the invariant points? I thought they had to be at 1/-1
• Feb 24th 2009, 10:28 AM
Reckoner
Here it is (click for bigger). Notice how the graph is vertically stretched, depending on the value of $\displaystyle a\text.$

Edit:

Quote:

Originally Posted by brentwoodbc
im trying to think of how you would find the invariant points.do you just go back to what the original equation was or?

like with 5/x what are the invariant points? I thought they had to be at 1/-1

There are no invariant points. For all nonzero $\displaystyle x\in\mathbb{R},\;\frac1x\neq\frac5x\text.$
• Feb 24th 2009, 10:42 AM
brentwoodbc
I chose the wrong word, I mean how would you find the points to graph it?

like with the reciprocal of the line y=x the "vertex" of 1/x (if thats what it is called with reciprocals) is at (1,1) right. so with 5/x what would that point be?
• Feb 24th 2009, 12:12 PM
Reckoner
Quote:

Originally Posted by brentwoodbc
I chose the wrong word, I mean how would you find the points to graph it?

like with the reciprocal of the line y=x the "vertex" of 1/x (if thats what it is called with reciprocals) is at (1,1) right. so with 5/x what would that point be?

I had typed up a response, but my browser crashed. Here:

I don't know if this is what you mean by "vertex," but usually the vertices of a curve are the points at which the derivative of its curvature is zero. The curvature $\displaystyle \kappa$ of $\displaystyle y=\frac5x,\,x>0$ is

$\displaystyle \kappa = \frac{|y''|}{(1+y'^2)^{3/2}}=\frac{|10/x^3|}{(1+25/x^4)^{3/2}}$

$\displaystyle =\frac{10|x^3|}{(x^4+25)^{3/2}}\text.$

Differentiating,

$\displaystyle \kappa'=\frac{30x^5(x^4+25)^{3/2}/|x^3|-60x^3|x^3|(x^4+25)^{1/2}}{(x^4+25)^3}$

$\displaystyle =\frac{30x^5(x^4+25)-60x^9}{|x^3|(x^4+25)^{5/2}}$

$\displaystyle =\frac{30x^5(25-x^4)}{|x^3|(x^4+25)^{5/2}}\text.$

So

$\displaystyle \kappa'=0\Rightarrow x=\pm\sqrt5$

and the vertices are at $\displaystyle \left(-\sqrt5,\,-\sqrt5\right)\text.$ and $\displaystyle \left(\sqrt5,\,\sqrt5\right)\text.$

In general, the vertices of $\displaystyle y=\frac ax,\,a\neq0$ should be at $\displaystyle \left(-\sqrt a,\,-\sqrt a\right)$ and $\displaystyle \left(\sqrt a,\,\sqrt a\right)\text.$

Edit: I wasn't really thinking of this when I did the above, but the points along the axis of symmetry will be vertices, and $\displaystyle \color{red}y = \frac ax$ is symmetric about the line $\displaystyle \color{red}y=x\text.$ So you could just find the points of intersection:

$\displaystyle \color{red}y=\frac ax\Rightarrow x=\frac ax\Rightarrow x^2=a\Rightarrow x=\pm\sqrt a$

Much simpler!