The problem in the image asks to graph a parabola with the following characteristics:

• x-intercepts at $x = -3$ and $x = 5$,
• minimum value at $y = -4$.

Step-by-step solution:

1. General form of a parabola: Since the parabola has x-intercepts at $x = -3$ and $x = 5$, its equation can be written in factored form: $y = a(x + 3)(x - 5)$ Here, $a$ is a constant that controls the "stretch" or "compression" of the parabola.

2. Use the minimum value: The minimum point (vertex) is located at the midpoint of the x-intercepts. The midpoint between $x = -3$ and $x = 5$ is: $x_{\text{vertex}} = \frac{-3 + 5}{2} = 1$ The minimum value of the parabola is given as $y = -4$, so the vertex is at $(1, -4)$.

3. Find the value of $a$: We can substitute the vertex $(1, -4)$ into the equation $y = a(x + 3)(x - 5)$ to find $a$. Substituting $x = 1$ and $y = -4$: $-4 = a(1 + 3)(1 - 5)$ $-4 = a(4)(-4)$ $-4 = -16a$ $a = \frac{-4}{-16} = \frac{1}{4}$ Thus, the equation of the parabola is: $y = \frac{1}{4}(x + 3)(x - 5)$

4. Final equation: Expanding this equation gives: $y = \frac{1}{4}(x^2 - 2x - 15)$ $y = \frac{1}{4}x^2 - \frac{1}{2}x - \frac{15}{4}$

The graph you've drawn corresponds to this equation. The parabola crosses the x-axis at $x = -3$ and $x = 5$, and it reaches its minimum value at $y = -4$ at $x = 1$.

Would you like further details or have any other questions?

Additional Questions:

1. How do you find the vertex of a parabola from its factored form?
2. Why does the parabola open upwards in this case?
3. Can you explain how to derive the general form from the factored form?
4. What does the constant $a$ represent in the parabola's equation?
5. How would the graph change if the constant $a$ was negative?

Tip:

Always verify the vertex and intercepts by plugging them into the equation, especially after determining the value of $a$.