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Math Help - Equation of a tangent to a curve calculus help!

  1. #1
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    Equation of a tangent to a curve calculus help!

    So for the following question, I have gone through the steps to solve it. I am just not confident in my answer so if anyone could help that would be appreciated!

    Q. Find the equation of the tangent to y = x^2 - 3x - 4 that is parallel to the line y = 7x + 3.

    A.
    1. The slope of y=7x+3 is 7, hence the slope of the tangent is 7.

    2. Finding the point on the curve where the derivative is 7.
    y'= 2x - 3
    7= 2x -3
    7-3 =2x
    4=2x
    x =2

    3. Solving for y;
    y= (2)^2 - 3(2) - 4
    y = -6

    4. The point on the curve is (2, -6)

    5. Using the coordinates to find the y-intercept:
    y= 7x + b
    -6 = 7(2) + b
    b= -20

    6. The equation of the tangent is y = 7x - 20


    Am I way off, or at least on the right track??? Thank you!
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  2. #2
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    Quote Originally Posted by kimberlyann View Post
    So for the following question, I have gone through the steps to solve it. I am just not confident in my answer so if anyone could help that would be appreciated!

    Q. Find the equation of the tangent to y = x^2 - 3x - 4 that is parallel to the line y = 7x + 3.

    A.
    1. The slope of y=7x+3 is 7, hence the slope of the tangent is 7.

    2. Finding the point on the curve where the derivative is 7.
    y'= 2x - 3
    7= 2x -3
    7-3 =2x correction ... 7+3 = 2x
    4=2x
    x =2

    3. Solving for y;
    y= (2)^2 - 3(2) - 4
    y = -6

    4. The point on the curve is (2, -6)

    5. Using the coordinates to find the y-intercept:
    y= 7x + b
    -6 = 7(2) + b
    b= -20

    6. The equation of the tangent is y = 7x - 20


    Am I way off, or at least on the right track??? Thank you!
    way off because of that simple error in algebra at the beginning.
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