Pre-Calculus

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Im having trouble with this.
Find an equation of the tangent line to the parabola at the given point and find the x-intercept of the line.
x^2=2y (4,8)

  • Pre-Calculus -

    y = (1/2) x^2
    The slope of the curve is dy/dx = x
    At x = 4, the slope is 4.
    For a line though (4,8) with slope = 4, the equation is
    (y-8) = 4(x-4)
    Rewrite that in standard y = mx + b form.

  • Pre-Calculus -

    well, from the above equation y = (x^2)/2

    to get an equation of a line you need a slope and a point. You have a point and to get the slope we need to get dy/dx
    and sub in the value for x in the point given [in this example 4]

    Then sub your point (x1,y1) and the slope (m) into
    y - y1 = m(x - x1)
    and then just tidy up

    Hope that helps

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