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Math Help - Equation of a Line Tangent

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    Equation of a Line Tangent

    How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?

    If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dumplings View Post
    How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?
    first find the slope at the point, that is, find f'(3)

    use that as your m in the point-slope form

    the tangent line will be given by: y - y_1 = m(x - x_1)

    where m is the slope and (x_1,y_1) is a point the line passes through, here, this is namely (3,14)


    If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?
    here m = -1 and (x_1,y_1) = (2,3) (do you see why?)

    use the same method as above
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by dumplings View Post
    How would you find an equation of the line tangent to f(x) = 2x^2 - 4 at teh point (3, 14) ?

    If f(2) = 3 and f(prime)(2) = -1, what would be the tangent line when x = 2?
    Finding an equation tangent to that line at (3, 14) involves finding f'(3) = 4(3) \rightarrow 12 The values of the derivative function are the slope values at those points. You have the slope, and the x and y coordinates, so therefore you can solve for the equation of the line:

    y=mx+b
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