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    Topics || Problems

    Find the slope of the curve \(y = (x+1)(x+2)\) at points where it crosses the x-axis. Trace the curve.

    When the curve crosses the \(x-axis\) the value of \(y = 0\).

    If \(y = 0\) then \( 0= x^2 +3x +2 \)

    \(x_1 = \frac{-3+\sqrt{9-4(2)}}{2} = -1\)

    \(x_2 = \frac{-3-\sqrt{9-4(2)}}{2} = -2\)

    \( y = (x+1)(x+2)\)

    \(\frac{dy}{dx} = (x+1) +(x+2)\)

    At \(x_1 = -1\)

    \(\frac{dy}{dx} = 0+1 = 1\)

    At \(x_1 = -2\)

    \(\frac{dy}{dx} =-1+0 = -1\)

    Thus the slope of the curve at \((-1,0) \) is 1 and -1 at \((-2,0) \)

    at points where it crosses the