Finding Equations of Tangents and Intersection Points for y=(2x-1)(x+1)

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additional maths help please!

My minds gone blank. How do I solve this
Find the equations of the tangents to the curve y=(2x-1)(x+1) at the points where the curve cuts the x axis. Find the point of intersection of these tangents.
Thanks
Emmab
 
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First find the tangent equation using differentiation, then find the points where the curve cuts the x-axis and plug in.
 


I would be inclined to do it the other way around! First find the x-values where the curve crosses the x-axis (y= 0 there), then find the derivative at those values of x. No real difference, of course.
 

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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