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    intersections

    Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chibiusagi View Post
    Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
    we can find how many zeros the function has by simply solving for them.

    set (x - 2) \left( x^2 + 2x + 6 \right) = 0

    \Rightarrow x - 2 = 0 or x^2 + 2x + 6 = 0

    \Rightarrow x = 2 or x = \frac {-2 \pm \sqrt{-20}}2

    clearly the quadratic has no real roots, thus the only root is x = 2

    to find the tangent line, use the point-slope form.

    y - y_1 = m(x - x_1)

    here, m = f'(2) and (x_1,y_1) = (2,0)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chibiusagi View Post
    Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
    y = (x - 2)(x^2 + 2x + 6)

    When this crosses the x-axis we have y = 0. So to find these x values:
    (x - 2)(x^2 + 2x + 6) = 0

    So either
    x - 2 = 0 \implies x = 2
    or
    x^2 + 2x + 6 = 0

    Note, however, that the quadratic factor here has no real zeros. So no real zeros are obtained by solving this equation.

    Thus the curve only crosses the x-axis once.

    Now to find the equation of the tangent.

    y^{\prime}(x) = 3x^2 + 2

    So at x = 2 the slope of the tangent to the curve is y^{\prime}(2) = 14.

    So we need the equation of a line with a slope of 14 that passes through the point (2, 0).

    y = 14x + b

    0 = 14 \cdot 2 + b \implies b = -28

    So the tangent line at (2, 0) is y = 14x - 28.

    -Dan
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