What Points on y=1/x Have a Tangent Line Slope of -4/9?

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I'm royally confused... don't even know where to start..

Find the point(s) on the graph of y = 1/x where the tangent line is parallel to 4x + 9y = 3. The m of 4x + 9y = 3 is -4/9 and sadly, that's all i know...
 
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can someone please help me?
 
joshb110 said:
can someone please help me?
First of all, what is the derivative of 1/x. Do you know how to calculate this? Remember that the derivative at some point x0 is the slope of the graph at that point.
 
hi. the derivative of 1/x = -1/ x^2, i believe.
 
what do i do next? do i just graph the two equations?
 
You mean as opposed to actually answering the question??

The problem, as you posted it, "Find the point(s) on the graph of y = 1/x where the tangent line is parallel to 4x + 9y = 3."

You know that the slope of 4x+ 9y= 3 is -4/9, you know that the derivative of y= 1/x at any x is y'= -1/x2 and you know, because apmcavoy just told you, that the derivative at a point is the slope of the tangent line there. For what x is the derivative equal to -4/9?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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