The graph provided represents a quadratic function, and we can derive its equation from key points. From the graph, the roots (x-intercepts) are $x = 2$ and $x = 10$, and the vertex is at $(6, -4)$.

The standard form of a quadratic function is:

$y = ax^2 + bx + c$

However, we will use the factored form first, based on the roots:

$y = a(x - 2)(x - 10)$

Now, we need to determine the value of $a$. Since the vertex is $(6, -4)$, we can substitute $x = 6$ and $y = -4$ into the equation:

$-4 = a(6 - 2)(6 - 10)$

Simplifying:

$-4 = a(4)(-4)$

$-4 = -16a$

Solving for $a$:

$a = \frac{-4}{-16} = \frac{1}{4}$

Thus, the equation of the quadratic function in factored form is:

$y = \frac{1}{4}(x - 2)(x - 10)$

Now, expand this to write it in standard form:

$y = \frac{1}{4}[(x - 2)(x - 10)] = \frac{1}{4}(x^2 - 12x + 20)$

Distribute the $\frac{1}{4}$:

$y = \frac{1}{4}x^2 - 3x + 5$

Therefore, the equation of the quadratic function in standard form is:

$y = \frac{1}{4}x^2 - 3x + 5$

Would you like further details or have any questions? Here are five related questions you might explore:

1. How can you find the vertex of a quadratic equation from its standard form?
2. What is the significance of the value of $a$ in the quadratic equation?
3. How would the graph change if $a$ was negative?
4. How do you convert a quadratic equation from factored form to standard form?
5. What methods can you use to find the roots of a quadratic equation?

Tip: Remember that the sign of $a$ determines the direction of the parabola—positive opens upwards, negative opens downwards!