We are given a graph of a quadratic function (a parabola) that passes through the point $(3, -23.5)$. We need to find the vertex of the parabola and write the equation of the parabola in vertex form.

1. Problem Description

We are given a graph of a quadratic function (a parabola) that passes through the point $(3, -23.5)$. We need to find the vertex of the parabola and write the equation of the parabola in vertex form.

2. Solution Steps

Step 1: Identify the vertex from the graph.

From the graph, the vertex appears to be at $(-1, 4)$.

Step 2: Write the vertex form of the quadratic equation.

The vertex form of a quadratic equation is given by:

$y = a(x - h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

Step 3: Substitute the vertex into the vertex form.

Substituting the vertex $(-1, 4)$ into the vertex form gives:

$y = a(x - (-1))^2 + 4$

$y = a(x + 1)^2 + 4$

Step 4: Use the given point to find the value of $a$.

We are given that the parabola passes through the point $(3, -23.5)$. Substitute $x = 3$ and $y = -23.5$ into the equation:

$-23.5 = a(3 + 1)^2 + 4$

$-23.5 = a(4)^2 + 4$

$-23.5 = 16a + 4$

Step 5: Solve for $a$.

Subtract 4 from both sides:

$-27.5 = 16a$

Divide both sides by 16:

$a = -\frac{27.5}{16} = -\frac{55}{32}$

Step 6: Write the equation of the parabola in vertex form.

Substitute the value of $a$ back into the vertex form equation:

$y = -\frac{55}{32}(x + 1)^2 + 4$

3. Final Answer

Vertex of the parabola is $(-1, 4)$.

The equation of the parabola in vertex form is $y = -\frac{55}{32}(x + 1)^2 + 4$.