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Mathematics 17 Online
OpenStudy (anonymous):

8.The straight line y=mx-1 is a tangent to the curve y=x^2. Find the possible values of m.

hartnn (hartnn):

Slope of tangent to a curve at a point is the derivative of the curve at that point.

hartnn (hartnn):

find dy/dx first..

OpenStudy (anonymous):

dy/dx ?

hartnn (hartnn):

haven't learnt derivatives yet? in calculus..

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

do u have any idea? @hartnn

OpenStudy (anonymous):

the derivative of x^2 is 2x, you should be able to go from there

hartnn (hartnn):

ok, then try to solve the equations simultaneously

hartnn (hartnn):

there should be only one point of intersection for the line to be a tangent

hartnn (hartnn):

y =x^2 y= mx -1 so x^2 = mx-1 got this ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay i got it

hartnn (hartnn):

good now the quadratic equation \(x^2- mx+1 =0\) has ONLY ONE root

OpenStudy (anonymous):

m=\[\pm2\]

hartnn (hartnn):

so its, \(b^2-4ac= 0 \)

hartnn (hartnn):

m^2 =4 and m = + or - 2 thats correct! good work!

OpenStudy (anonymous):

thnx a lot. :] @hartnn

OpenStudy (anonymous):

and @msasu25

hartnn (hartnn):

welcome ^_^

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