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Math Help - Finding the "normal line" of a graph.

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    Finding the "normal line" of a graph.

    Here's the problem:

    "Given the parabola y = x^2 - Ax + B, where A is the number of letters in your first name and B is the number of letters in your last name, find an equation for the line L that contains the point (0,B) and is perpendicular to the tangent line to the parabola at (0,B). This is called the 'normal line' to the graph."

    For me, A = 8 and B = 6. It says that I can find the slope of the tangent line and then use that to compute the slope of the normal line, but I'm not sure how to go about doing that. I know the derivative of the function is y = 2x + A (or is it y = 2x - A?), but I don't know where to go from there.
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  2. #2
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    Is there no one that can help me with this? D:
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  3. #3
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    Quote Originally Posted by Rumor View Post
    Here's the problem:

    "Given the parabola y = x^2 - Ax + B, where A is the number of letters in your first name and B is the number of letters in your last name, find an equation for the line L that contains the point (0,B) and is perpendicular to the tangent line to the parabola at (0,B). This is called the 'normal line' to the graph."

    For me, A = 8 and B = 6. It says that I can find the slope of the tangent line and then use that to compute the slope of the normal line, but I'm not sure how to go about doing that. I know the derivative of the function is y = 2x + A (or is it y = 2x - A?), but I don't know where to go from there.
    \frac{d}{dx}(x^2-Ax+B) = 2x-A

    Compare to the equation of a line: y = mx+c

    For the normal line recall the product of perpendicular lines is -1
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