The graph represents a quadratic curve, and you are asked to find the equation of the curve in the form:

$y = ax^2 + bx + c$

We are given the following key points from the graph:

1. The x-intercepts are $x = -1$ and $x = 3$, which gives the roots of the quadratic equation.
2. The vertex is at $(1, -9)$.

Step 1: Form the general equation

Since the quadratic equation has roots at $x = -1$ and $x = 3$, we can write it in its factored form as:

$y = a(x + 1)(x - 3)$

Here, $a$ is a constant that needs to be determined.

Step 2: Use the vertex to find $a$

The vertex of the parabola is given as $(1, -9)$, meaning when $x = 1$, $y = -9$. We can substitute $x = 1$ and $y = -9$ into the equation to find the value of $a$.

$-9 = a(1 + 1)(1 - 3)$ $-9 = a(2)(-2)$ $-9 = -4a$ $a = \frac{-9}{-4} = \frac{9}{4}$

Thus, the equation of the quadratic becomes:

$y = \frac{9}{4}(x + 1)(x - 3)$

Step 3: Expand the equation

Now, expand the equation to get it in the form $y = ax^2 + bx + c$.

$y = \frac{9}{4}(x^2 - 3x + x - 3)$ $y = \frac{9}{4}(x^2 - 2x - 3)$ $y = \frac{9}{4}x^2 - \frac{18}{4}x - \frac{27}{4}$

Finally, we multiply through by 4 to clear the fractions:

$4y = 9x^2 - 18x - 27$

So, the equation of the curve is:

$y = \frac{9}{4}x^2 - \frac{18}{4}x - \frac{27}{4}$