1. **Problem Statement:** Show using the First Derivative Test that for the quadratic function $y = ax^2 + bx + c$ with $a \neq 0$, the graph has a relative maximum at the vertex if $a < 0$ and a relative minimum at the vertex if $a > 0$. 2. **Formula and Important Rules:** The first derivative of $y$ is given by: $$y' = \frac{dy}{dx} = 2ax + b$$ The vertex occurs where $y' = 0$, so: $$2ax + b = 0 \implies x = -\frac{b}{2a}$$ The First Derivative Test states: - If $y'$ changes from positive to negative at a critical point, the function has a relative maximum there. - If $y'$ changes from negative to positive at a critical point, the function has a relative minimum there. 3. **Find the critical point (vertex):** Set $y' = 0$: $$2ax + b = 0 \implies x = -\frac{b}{2a}$$ 4. **Analyze the sign of $y'$ around the vertex:** - For $x < -\frac{b}{2a}$: $$y' = 2a x + b$$ Since $x$ is less than the vertex, consider $x = -\frac{b}{2a} - h$ for small $h > 0$: $$y' = 2a\left(-\frac{b}{2a} - h\right) + b = -b - 2ah + b = -2ah$$ - For $x > -\frac{b}{2a}$: $$y' = 2a\left(-\frac{b}{2a} + h\right) + b = -b + 2ah + b = 2ah$$ 5. **Sign of $y'$ depends on $a$:** - If $a > 0$: - For $x < -\frac{b}{2a}$, $y' = -2ah < 0$ (negative) - For $x > -\frac{b}{2a}$, $y' = 2ah > 0$ (positive) - So $y'$ changes from negative to positive, indicating a relative minimum at the vertex. - If $a < 0$: - For $x < -\frac{b}{2a}$, $y' = -2ah > 0$ (positive) - For $x > -\frac{b}{2a}$, $y' = 2ah < 0$ (negative) - So $y'$ changes from positive to negative, indicating a relative maximum at the vertex. 6. **Conclusion:** Using the First Derivative Test, the quadratic function $y = ax^2 + bx + c$ has: - A relative minimum at the vertex when $a > 0$. - A relative maximum at the vertex when $a < 0$. This matches the known shape of parabolas opening upwards ($a > 0$) and downwards ($a < 0$).