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

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    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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  2. #2
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    Quote Originally Posted by plstevens View Post
    y=3x^2-x^3

    P= (1,2)

    y=
    The general equation of a line is y - y_1 = m(x - x_1) where m is the gradient and (x_1, y_1) is a known point on the line.

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

    To get m, find \frac{dy}{dx} of y = 3x^2 - x^3 and evaluate at x = 1.
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    I still don't understand
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    Quote Originally Posted by plstevens View Post
    I still don't understand
    What exactly don't you understand?

    I've said that in <br />
y - y_1 = m(x - x_1)<br />
, <br />
(x_1, y_1)<br />
is a known point.

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

    That means you substitute x_1 = 1 and y_1 = 2 into <br />
y - y_1 = m(x - x_1)<br />
.

    Now get the value of m. How I said to in my first reply. Then substitute the value into <br />
y - y_1 = m(x - x_1)<br />
.
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    is it y=3x-1
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    Quote Originally Posted by plstevens View Post
    is it y=3x-1
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