The task in the image is to find the equation of the parabola shown in the figure, in the form:

$y = a(x - x_1)(x - x_2)$

where $(x_1, 0)$ and $(x_2, 0)$ are the x-intercepts of the parabola. From the graph, it is clear that the parabola opens upwards and passes through the point $(4, -4)$. We can proceed with the following steps:

Step 1: Identify the x-intercepts

From the graph, it seems that the parabola touches the x-axis at $x_1 = -4$ and $x_2 = 4$. These are the roots of the parabola.

Step 2: Use the general form $y = a(x - x_1)(x - x_2)$

Substitute $x_1 = -4$ and $x_2 = 4$ into the equation:

$y = a(x + 4)(x - 4)$

This simplifies to:

$y = a(x^2 - 16)$

Step 3: Find the value of $a$

To determine $a$, use the fact that the parabola passes through the point $(0, -4)$. Substitute $x = 0$ and $y = -4$ into the equation:

$-4 = a(0^2 - 16)$ $-4 = -16a$ $a = \frac{-4}{-16} = \frac{1}{4}$

Step 4: Write the final equation

Substitute $a = \frac{1}{4}$ into the equation:

$y = \frac{1}{4}(x^2 - 16)$

Thus, the equation of the parabola is:

$y = \frac{1}{4}x^2 - 4$

Let me know if you need further details or clarifications!

Related Questions:

1. How do you determine the vertex of a parabola given its equation in standard form?
2. What is the relationship between the coefficient $a$ and the direction in which the parabola opens?
3. How do you find the focus of a parabola given its equation?
4. What are the properties of the parabola when $a > 0$ compared to when $a < 0$?
5. How do the x-intercepts help in constructing the factored form of a quadratic equation?

Tip:

To easily identify the shape of a parabola, remember that when $a > 0$, the parabola opens upwards, and when $a < 0$, it opens downwards.