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= 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
Originally Posted by plstevens I still don't understand 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
Originally Posted by plstevens is it y=3x-1
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