This is a question about determining the graph of a quadratic function $y = ax^2 + bx + c$ based on the given conditions:

1. $a < 0$: The coefficient of $x^2$ is negative, so the parabola opens downwards.
2. $b < 0$: The coefficient of $x$ affects the symmetry and position of the vertex, but it doesn't directly determine whether the parabola opens up or down.
3. $c > 0$: The constant $c$ is positive, so the $y$-intercept is above the $x$-axis.

Based on these conditions:

• The parabola opens downwards ($a < 0$).
• The $y$-intercept is above the $x$-axis ($c > 0$).
• The vertex will likely be in the left half-plane ($b < 0$).

From the given graphs, the correct choice is D, as it is the only one with a downward-opening parabola with a $y$-intercept above the $x$-axis.

Let me know if you'd like further explanation! Here are some related questions:

1. What happens to the graph if $a > 0$ but $c > 0$?
2. How does $b$ influence the symmetry of the parabola?
3. Can a quadratic function have more than one $y$-intercept? Why or why not?
4. What is the equation of the axis of symmetry for a quadratic function?
5. How does the discriminant $\Delta = b^2 - 4ac$ affect the shape of the parabola?

Tip: The sign of $a$ always determines the opening direction of the parabola (upwards if $a > 0$, downwards if $a < 0$).