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?
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!