TANGENT???????????????????

The slope of the tangent line to the parabola y = 3 x^2 + 4 x + 2 at the point ( 2 , 22 ) is: 2

The equation of this tangent line can be written in the form y = mx+b where m is:_______________?

and where b is:_______________?

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  • 8 years ago
    Best Answer

    Hi Fapre,

    Agree with Iobo that m is the slope of the tangent line and b is the y-intercept of the tangent line. However, I would just add what the actual values are when applied to the parabola and its tangent you cited.

    In this case, y' = 6x + 4 means that y'(2) = 16. So the the slope of the tangent line at x=2 is m=16.

    To find b, the y-intercept of the tangent line, just use y=mx+b, your point (2,22) as (x,y), and m=16:

    That gives 22 = 16(2) + b, or b = -10. Thus the equation of the tangent line to the parabola is y = 16x - 10.

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  • Anonymous
    8 years ago

    First, find the derivative:

    dy/dx=6x+4

    Now, plug in x:

    dy/dx=6(2)+4=16<------this is your slope (m)

    Now plug this into the y=mx+b equation given your x,y coordinates to solve for b:

    22=16(2)+b

    22=32+b

    b=-10

    So:

    m is 16

    b is -10

    And your equation is y=16x-10

  • 8 years ago

    y = mx + b is known as the "slope/intercept" form of a line equation.

    m is the slope of the line and b is the y intercept of the line.

    You plug in the 3 known quantities x, y and m (all are given to you) and solve for b.

  • 8 years ago

    y = 3x² + 4x + 2

    dy/dx = 6x + 4

    x = 2

    y = 22

    dy/dx = 6(2) + 4

    dy/dx = 12 + 4

    dy/dx = 16

    b = y - [(dy/dx)(x)]

    b = 22 - [16(2)

    b = 22 - 32

    b = - 10

    Equation of Tangent Line:

    y = 16x - 10

    ¯¯¯¯¯¯¯¯¯¯¯

    m = 16

    ¯¯¯¯¯¯¯

    b = - 10

    ¯¯¯¯¯¯¯

     

    Source(s): 5/9/12
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  • lobo
    Lv 6
    8 years ago

    m is slope of the tangent line

    b is the y intercept of the tangent line.

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