For example, lets say the function of the curve is y = x^3

How would you find two lines that pass through points (2, 8) and are tangent to the curve?

Many thanks.

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- May 13th 2011, 06:33 PMsamstarkHow do I find two lines that pass through points (x, y) and are tangent to the curve?
For example, lets say the function of the curve is y = x^3

How would you find two lines that pass through points (2, 8) and are tangent to the curve?

Many thanks. - May 13th 2011, 06:39 PMProve It
It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...

Your tangent is a line, so is of the form y = mx + c.

The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.

So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.

Can you get to there? Can you see what to do next? - May 13th 2011, 06:53 PMsamstark
- May 13th 2011, 06:56 PMProve It
OK, you don't know the point where the tangent will touch your curve, so you are just going to use your derivative as your m value for the moment.

Substituting gives y = (3x^2)x + c, or y = 3x^3 + c

Now use your point to find c... - May 13th 2011, 07:05 PMsamstark
Am I supposed to substitute this way?

8 = 3(2)^3 + c

8 = 24 + c

c = 8 - 24 or c = -16 - May 13th 2011, 07:07 PMProve It
Correct.

That means your tangent line is of the form y = mx - 16.

You are almost there, now you need to use your point to find m. - May 13th 2011, 07:13 PMsamstark
- May 13th 2011, 07:20 PMProve It
Looks good to me...

- May 13th 2011, 07:26 PMsamstark
Thanks a lot! :D