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8.The straight line y=mx-1 is a tangent to the curve y=x^2. Find the possible values of m.
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Slope of tangent to a curve at a point is the derivative of the curve at that point.
find dy/dx first..
dy/dx ?
haven't learnt derivatives yet? in calculus..
nope
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do u have any idea? @hartnn
the derivative of x^2 is 2x, you should be able to go from there
ok, then try to solve the equations simultaneously
there should be only one point of intersection for the line to be a tangent
y =x^2 y= mx -1 so x^2 = mx-1 got this ?
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yes
okay i got it
good now the quadratic equation \(x^2- mx+1 =0\) has ONLY ONE root
m=\[\pm2\]
so its, \(b^2-4ac= 0 \)
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m^2 =4 and m = + or - 2 thats correct! good work!
thnx a lot. :] @hartnn
and @msasu25
welcome ^_^
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