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Math Help - equation of tangent

  1. #1
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    equation of tangent

    The normal to the curve y=x^2 - 6x + 1 at the point p is parallel to the straight line.Find the equation of the 2y = x + 3 tangent to the curve at the point p
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    The normal to the curve y=x^2 - 6x + 1 at the point p is parallel to the straight line 2y = x + 3.

    Find the equation of the tangent to the curve at the point p
    If this is how your post should have read, then calculating the derivative of x^2-6x+1

    will give you the slope of the tangent.

    The slope of the line 2y=x+3\Rightarrow\ y=\frac{1}{2}x+\frac{3}{2}
    is \frac{1}{2}

    The normal and the tangent are perpendicular, so you can find what the tangent slope needs to be.

    The derivative being equal to this tangent slope will obtain the x co-ordinate of p

    The y co-ordinate is obtained by placing the x co-ordinate into f(x).

    Finally, to get the equation of the tangent, use the equation of the line formula,
    using the slope calculated and the co-ordinates of p
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  3. #3
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    i get:

    y=x^2 -3x + 5

    \frac{dy}{dx} = 2x - 6

    2y = x + 3

    y = \frac{1}{2}x + \frac{3}{2}

    (m_1)(m_2) = -1

    \frac{1}{2}(m_2) = -1

    m_2 = -2

    2x - 6 = -2

    x = 2

    when x = 2

    y = 2^2 - 6(2) + 1

    y = -7

    using (2,-7) and m = -2

    y + 7 = -2 (x - 2)

    y + 7 = -2x + 4

    is that correct sir?
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  4. #4
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    Quote Originally Posted by mastermin346 View Post
    i get:

    y=x^2 -6x + 1

    \frac{dy}{dx} = 2x - 6

    2y = x + 3

    y = \frac{1}{2}x + \frac{3}{2}

    (m_1)(m_2) = -1

    \frac{1}{2}(m_2) = -1

    m_2 = -2

    2x - 6 = -2

    x = 2

    when x = 2

    y = 2^2 - 6(2) + 1

    y = -7

    using (2,-7) and m = -2

    y + 7 = -2 (x - 2)

    y + 7 = -2x + 4

    is that correct sir?
    Yes, that's it. (small typo on your first line)
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  5. #5
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    Quote Originally Posted by mastermin346 View Post
    The normal to the curve y=x^2 - 6x + 1 at the point p is parallel to the straight line.Find the equation of the 2y = x + 3 tangent to the curve at the point p
    Shouldn't it say the following?

    The normal to the curve y=x^2 - 6x + 1 at the point p is parallel to the straight line 2y = x + 3. Find the equation of the tangent to the curve at the point p.

    Did you copy-paste from a two-column PDF file?
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