1. **State the problem:** Find the gradient of the curve $$y = x - \frac{3}{x+2}$$ at the points where the curve crosses the $$x$$-axis. 2. **Find the points where the curve crosses the $$x$$-axis:** This means solving for $$x$$ when $$y=0$$. $$0 = x - \frac{3}{x+2}$$ Multiply both sides by $$x+2$$ to clear the denominator: $$0 = x(x+2) - 3$$ $$0 = x^2 + 2x - 3$$ 3. **Solve the quadratic equation:** $$x^2 + 2x - 3 = 0$$ Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=2$$, $$c=-3$$: $$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2}$$ $$x = \frac{-2 \pm 4}{2}$$ So the two solutions are: $$x = \frac{-2 + 4}{2} = 1$$ $$x = \frac{-2 - 4}{2} = -3$$ 4. **Find the gradient function by differentiating $$y$$ with respect to $$x$$:** $$y = x - \frac{3}{x+2} = x - 3(x+2)^{-1}$$ Differentiate term by term: $$\frac{dy}{dx} = 1 - 3 \cdot (-1)(x+2)^{-2} = 1 + \frac{3}{(x+2)^2}$$ 5. **Evaluate the gradient at the points where $$y=0$$:** At $$x=1$$: $$\frac{dy}{dx} = 1 + \frac{3}{(1+2)^2} = 1 + \frac{3}{3^2} = 1 + \frac{3}{9} = 1 + \frac{1}{3} = \frac{4}{3}$$ At $$x=-3$$: $$\frac{dy}{dx} = 1 + \frac{3}{(-3+2)^2} = 1 + \frac{3}{(-1)^2} = 1 + 3 = 4$$ **Final answer:** The gradient of the curve at the points where it crosses the $$x$$-axis are: - At $$x=1$$, gradient $$= \frac{4}{3}$$ - At $$x=-3$$, gradient $$= 4$$