Struggling to Graph with Respect to y? Tips and Tricks for TI-84 Users

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Homework Statement


I'm having difficulty graphing with respect to y (unless I graph it point by point).

For example, I find it hard to graph: f(y)= \frac{y}{\sqrt{16-y^2}}


Does anyone know of an easy method to graph such a function or know how to do it on a Ti 84+ SE? Also, it would be great if someone could link me to a website that shows you a method. Thanks!
 
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Whenever you graph something, you need to first identify the independant and dependant variables. In this case, y is the independant variable and f(y) is the dependant variable so it is really no different than graphing an equation with x and f(x).

Most graphing calculators will only let you call the independant variable "x" and the dependant variable "y" so this is a time when you just have to be smarter than the calculator.
 
Graph y= f(x)= \frac{x}{\sqrt{16-x^2}}
and then turn your calculator on its side!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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