Find the equation of the tangents to the curve y=(2#x-1)(#x-1)
# means square root btw.
at the points were the curve cuts the x-axis (y=0)
Find the co-ordinates of the points of intersection of these tangents.
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- Thread Starter
- 24-01-2010 20:04
- 24-01-2010 20:12
okay, first find the x values. so you have 2#x = 1 and #x = 1. so x = 1 and (1/#2) .
then find dy/dx, substitute these x values in to find the gradient. then do the usual y=mx + c by putting in the x and y to find c so you then have the eqn's of the two tangents.
then when you've done that, make them equal to each other to find the x value of where they intersect, sub it into one of the tangent eqn's to find y. and you're done.