Math Help - [SOLVED] Graph question

1. [SOLVED] Graph question

whats the difference between say...

$5/x$ or $1/x$ 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?

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

$5/x$ or $1/x$ 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 $y=\frac1x,$ then $\frac5x$ is just $5y\text.$ 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.

3. thanks

4. 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

5. Here it is (click for bigger). Notice how the graph is vertically stretched, depending on the value of $a\text.$

Edit:

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 $x\in\mathbb{R},\;\frac1x\neq\frac5x\text.$

6. 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?

7. 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 $\kappa$ of $y=\frac5x,\,x>0$ is

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

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

Differentiating,

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

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

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

So

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

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

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

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 $\color{red}y = \frac ax$ is symmetric about the line $\color{red}y=x\text.$ So you could just find the points of intersection:

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

Much simpler!