Let's analyze the conditions given and verify if $a < 0$, $b < 0$, and $c > 0$ can all be true.

Key Points:

1. Downward-opening parabola: For a parabola that opens downward, the coefficient $a$ must be negative ($a < 0$).

2. Vertex in the fourth quadrant: The vertex $(h, k)$ of the parabola is given by: $h = -\frac{b}{2a}, \quad k = c - \frac{b^2}{4a}$ For the vertex to be in the fourth quadrant, $h > 0$ and $k < 0$.

Now, let’s explore the inequalities for $b$ and $c$:

Condition for $b$:

For $h > 0$, we need $-\frac{b}{2a} > 0$. Since $a < 0$, $b$ must also be negative ($b < 0$) to make $h$ positive.

Condition for $c$:

For the vertex’s $y$-coordinate $k$ to be negative, we need $c - \frac{b^2}{4a} < 0$. Since $a < 0$ and $b < 0$, $\frac{b^2}{4a}$ is positive. Therefore, $c$ must be positive ($c > 0$) for $k$ to be negative.

Conclusion

Yes, it is possible for $a < 0$, $b < 0$, and $c > 0$ to all be true, and these conditions will indeed place the vertex of a downward-opening parabola in the fourth quadrant.

Would you like more details on any part of this solution?

Here are five related questions:

1. How does changing $c$ affect the position of the vertex?
2. What would happen if $a > 0$ with $b < 0$ and $c > 0$?
3. How can we find the intercepts of this parabola?
4. How does the value of $b$ impact the symmetry of the parabola?
5. What are the conditions for a parabola to have its vertex in each of the four quadrants?

Tip: The sign of $a$ in the quadratic equation $y = ax^2 + bx + c$ always determines the direction the parabola opens—negative for downward, positive for upward.