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Thread: Tangent value

  1. #1
    Forum Admin topsquark's Avatar
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    Tangent value

    This is a question by declon.
    Quote Originally Posted by declon
    Hi, would you be able to help me? I usually sit for hours trying to figure out problems but there's just some I can't figure out.

    Question:
    The line y = 4x + k is a tangent to the curve y = x^2 + 4x + 3 at a point P.
    (b) Find the value of k.
    (c) Find an equation of the normal to the curve y = x^2 + 4x + 3 at the point P.

    I made the line and curve equal so 4x + k = x^2 +4x + 3 then rearranged for x^2 = k - 3 but then I don't know where to go from there

    Regards
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  2. #2
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    Re: Tangent value

    Question:
    The line y = 4x + k is a tangent to the curve y = x^2 + 4x + 3 at a point P.
    (b) Find the value of k.
    (c) Find an equation of the normal to the curve y = x^2 + 4x + 3 at the point P.

    Ok, let's think this through. A diagram often helps your thinking.
    Draw a quick sketch of the parabola y = x^2 + 4x + 3. That is concave up and cuts y-axis at 3. Don't worry about x-intercepts.

    Now draw a line (upward sloping as y = 4x + k has a gradient of +4) which touches the parabola at P.

    Now the slope at P on the parabola is the same as the slope at P on the straight line. Yes? And that slope is 4. Right?

    Now the slope of the parabola is given by the derivative of x^2 + 4x + 3.

    This derivative is ................... and it is equal to ............ So you can work out the x-coordinate of P. Use it to find the y-coordinate of P.

    (b) Now find the equation of the tangent at P and hence find k.
    (c)You've got the point P, you can easily work out the slope of the normal, and use them to find the equation of the normal.
    Thanks from topsquark
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