The graph shows a quadratic function with intercepts at $(-2, 0)$ and $(6, 0)$, and the vertex is at $(4, -3)$. To write the quadratic function in standard form, $f(x) = ax^2 + bx + c$, we can follow these steps:

Step 1: Identify the factored form of the quadratic.

Since the x-intercepts are $x = -2$ and $x = 6$, the function in factored form is: $f(x) = a(x + 2)(x - 6)$ where $a$ is a constant to be determined.

Step 2: Use the vertex to find $a$.

We know the vertex is $(4, -3)$, so substitute $x = 4$ and $f(x) = -3$ into the equation to solve for $a$: $-3 = a(4 + 2)(4 - 6)$ $-3 = a(6)(-2)$ $-3 = -12a$ $a = \frac{-3}{-12} = \frac{1}{4}$

Step 3: Write the equation in factored form.

Now that we know $a = \frac{1}{4}$, the function in factored form is: $f(x) = \frac{1}{4}(x + 2)(x - 6)$

Step 4: Expand to write in standard form.

To convert this into standard form, expand the factors: $f(x) = \frac{1}{4}(x^2 - 6x + 2x - 12)$ $f(x) = \frac{1}{4}(x^2 - 4x - 12)$ $f(x) = \frac{1}{4}x^2 - \frac{1}{4}(4x) - \frac{1}{4}(12)$ $f(x) = \frac{1}{4}x^2 - x - 3$

Thus, the quadratic function in standard form is: $f(x) = \frac{1}{4}x^2 - x - 3$

Would you like more details or have any further questions?

Here are 5 related questions to explore further:

1. How do you convert a quadratic function from vertex form to standard form?
2. What is the significance of the vertex in a quadratic function?
3. How do you determine the direction in which a parabola opens?
4. How can you find the axis of symmetry of a parabola?
5. What methods can be used to find the roots of a quadratic equation?

Tip: Remember, the vertex form of a quadratic equation is particularly useful when you know the vertex, making it easier to graph.