Question

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

Answer 1

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

Answer 2

find dy/dx first..

Answer 3

dy/dx ?

Answer 4

haven't learnt derivatives yet? in calculus..

Answer 5

nope

Answer 6

do u have any idea? @hartnn

Answer 7

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

Answer 8

ok, then try to solve the equations simultaneously

Answer 9

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

Answer 10

y =x^2
y= mx -1

so
x^2 = mx-1
got this ?

Answer 11

yes

Answer 12

okay i got it

Answer 13

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

Answer 14

m=\[\pm2\]

Answer 15

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

Answer 16

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

Answer 17

thnx a lot. :] @hartnn

Answer 18

and @msasu25

Answer 19

welcome ^_^