Find the equation of the tangent line to the following curve at the given point P. Il

y=3x^2-x^3

P= (1,2)

y=

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Originally Posted by plstevens

y=3x^2-x^3

P= (1,2)

y=

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The general equation of a line is where m is the gradient and is a known point on the line.

In your question, a known point is (1, 2)

To get m, find of and evaluate at x = 1.

I still don't understand

Quote:

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Originally Posted by plstevens

I still don't understand

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What exactly don't you understand?

I've said that in , is a known point.

A known point is (1, 2). It's given to you on a platter.

That means you substitute and into .

Now get the value of m. How I said to in my first reply. Then substitute the value into .

is it y=3x-1

Quote:

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Originally Posted by plstevens

is it y=3x-1

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(Clapping)