1. **Problem a:** Find the equation of the tangent to the curve $$y=\sqrt{x+1}-a+x=4$$. 2. **Step 1:** Clarify the function. The equation seems ambiguous. Assuming the function is $$y=\sqrt{x+1}-a+x$$ and it equals 4 at some point. 3. **Step 2:** Find the derivative $$y'$$ to get the slope of the tangent. Given $$y=\sqrt{x+1}-a+x$$, rewrite as $$y=(x+1)^{1/2} - a + x$$. 4. **Step 3:** Differentiate: $$y' = \frac{1}{2\sqrt{x+1}} + 1$$ 5. **Step 4:** Find the point where $$y=4$$: $$\sqrt{x+1} - a + x = 4$$ Without a specific value for $$a$$, we cannot find exact $$x$$. 6. **Step 5:** The tangent line at point $$x_0$$ has slope $$m = y'(x_0)$$ and passes through $$\left(x_0, y(x_0)\right)$$. Equation of tangent: $$y - y(x_0) = m (x - x_0)$$ --- 7. **Problem b:** Find the equation of the normal to $$y = x^3 - 5x + 2$$ at $$x = -2$$. 8. **Step 1:** Find $$y(-2)$$: $$y(-2) = (-2)^3 - 5(-2) + 2 = -8 + 10 + 2 = 4$$ 9. **Step 2:** Find derivative $$y'$$: $$y' = 3x^2 - 5$$ 10. **Step 3:** Find slope of tangent at $$x=-2$$: $$m_{tangent} = 3(-2)^2 - 5 = 3(4) - 5 = 12 - 5 = 7$$ 11. **Step 4:** Slope of normal is negative reciprocal: $$m_{normal} = -\frac{1}{7}$$ 12. **Step 5:** Equation of normal line passing through $$(-2,4)$$: $$y - 4 = -\frac{1}{7}(x + 2)$$ 13. **Step 6:** Simplify: $$y = -\frac{1}{7}x - \frac{2}{7} + 4 = -\frac{1}{7}x + \frac{26}{7}$$ **Final answers:** - a) Tangent equation depends on $$a$$ and point $$x_0$$ satisfying $$\sqrt{x_0+1} - a + x_0 = 4$$ with slope $$m = \frac{1}{2\sqrt{x_0+1}} + 1$$. - b) Normal equation: $$y = -\frac{1}{7}x + \frac{26}{7}$$