1. **State the problem:** Find the equation of the normal to the function $f(x) = ax^2 + bx + c$ at the turning point. 2. **Recall the turning point:** The turning point of a quadratic function $f(x) = ax^2 + bx + c$ occurs where the derivative $f'(x) = 0$. 3. **Find the derivative:** $$f'(x) = 2ax + b$$ 4. **Set derivative to zero to find turning point $x$-coordinate:** $$0 = 2ax + b$$ $$2ax = -b$$ $$x = \frac{-b}{2a}$$ 5. **Find the $y$-coordinate of the turning point by substituting $x = \frac{-b}{2a}$ into $f(x)$:** $$y = a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c$$ $$= a\frac{b^2}{4a^2} - \frac{b^2}{2a} + c$$ $$= \frac{b^2}{4a} - \frac{b^2}{2a} + c$$ $$= \frac{b^2}{4a} - \frac{2b^2}{4a} + c$$ $$= -\frac{b^2}{4a} + c$$ 6. **Find the slope of the tangent at the turning point:** Since $f'(x) = 0$ at the turning point, the slope of the tangent is 0. 7. **Find the slope of the normal:** The normal is perpendicular to the tangent, so its slope is the negative reciprocal of 0, which is undefined (vertical line). 8. **Equation of the normal:** Since the normal is vertical at $x = \frac{-b}{2a}$, the equation is: $$x = \frac{-b}{2a}$$ **Answer:** Option 3) $x = \frac{-b}{2a}$