Parallel Tangents to Cubic Graph

flyinghigh
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Homework Statement


I have the graph of a function f(x)=(x+1)(x+4)(x+6). I've found the tangent at x=-6, the equation of which is y=10x+60. I then need to algebraically find the equation of another tangent on the curve which is parallel to the first.

Homework Equations


No idea.

The Attempt at a Solution


Since I'm graphing this on Autograph I've managed to find a point where the tangent is parallel to the first, but since this can be done without the computer program I'm really interested to know how it's done. So yeah, I know that the tangent I'm trying to find will have a gradient of 10 but that's about it...

Any help would be great,
flyinghigh
 
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Okay, you know you're looking for a tangent line with a slope of 10.

Can you write an equation for the slope of the tangent (as a function of x), and set that equal to 10?
 

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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