1. **Problem Statement:** Given the quadratic function $y = ax^2 + bx + c$ with conditions $a < 0$, $b > 0$, and $c < 0$, describe the graph's shape and position. 2. **Formula and Key Properties:** The quadratic function is $y = ax^2 + bx + c$. - Since $a < 0$, the parabola opens downward. - The $y$-intercept is at $(0, c)$, and since $c < 0$, it lies below the $x$-axis. - The slope at the $y$-intercept is given by the derivative $y' = 2ax + b$. At $x=0$, slope is $b > 0$, so the curve slopes upward (to the right) at the $y$-intercept. 3. **Vertex Calculation:** The vertex $x$-coordinate is given by $$x_v = -\frac{b}{2a}.$$ Since $a < 0$ and $b > 0$, $x_v = -\frac{b}{2a}$ is positive (because dividing a positive $b$ by a negative $2a$ gives a negative denominator, so the whole fraction is positive). 4. **Vertex $y$-coordinate:** Calculate vertex $y$-value: $$y_v = a x_v^2 + b x_v + c = a \left(-\frac{b}{2a}\right)^2 + b \left(-\frac{b}{2a}\right) + c.$$ Simplify: $$y_v = a \frac{b^2}{4a^2} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{2b^2}{4a} + c = -\frac{b^2}{4a} + c.$$ Since $a < 0$, $-\frac{b^2}{4a} > 0$, so $y_v$ is the sum of a positive number and $c$ (which is negative). The vertex $y$-value is above the $x$-axis if this sum is positive. 5. **Summary:** - Parabola opens downward. - Vertex is to the right of the $y$-axis ($x_v > 0$). - Vertex is above the $x$-axis. - $y$-intercept is below the $x$-axis. - Slope at $y$-intercept is positive, so the curve slopes upward at $x=0$. This matches the description of the parabola in the bottom-center-right position: opens downward, vertex above $x$-axis, crosses $y$-axis below zero. **Final answer:** The parabola opens downward with vertex above the $x$-axis to the right of the $y$-axis, and crosses the $y$-axis below zero with positive slope there.