1. **Problem Statement:** Find the equation of the tangent line to the curve at the given $x$ value. 2. **Formula and Rules:** The equation of the tangent line at $x=a$ is given by: $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of the function evaluated at $x=a$. --- ### a) For $y = 2x^2 + x - 2$ at $x=0$: 3. Find the derivative: $$f'(x) = \frac{d}{dx}(2x^2 + x - 2) = 4x + 1$$ 4. Evaluate the function and derivative at $x=0$: $$f(0) = 2(0)^2 + 0 - 2 = -2$$ $$f'(0) = 4(0) + 1 = 1$$ 5. Write the tangent line equation: $$y = f(0) + f'(0)(x - 0) = -2 + 1 \cdot x = x - 2$$ --- ### b) For $y = \sqrt{x} + 2$ at $x=1$: 6. Rewrite the function for differentiation: $$y = x^{1/2} + 2$$ 7. Find the derivative: $$f'(x) = \frac{d}{dx}(x^{1/2} + 2) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$ 8. Evaluate the function and derivative at $x=1$: $$f(1) = \sqrt{1} + 2 = 1 + 2 = 3$$ $$f'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2}$$ 9. Write the tangent line equation: $$y = f(1) + f'(1)(x - 1) = 3 + \frac{1}{2}(x - 1) = \frac{1}{2}x + \frac{5}{2}$$ --- **Final answers:** - a) $y = x - 2$ - b) $y = \frac{1}{2}x + \frac{5}{2}$