The problem asks you to write a quadratic function with zeros $-8$ and $-6$.

A quadratic function can be expressed in factored form as: $f(x) = a(x - r_1)(x - r_2)$ where $r_1$ and $r_2$ are the zeros of the function. In this case, $r_1 = -8$ and $r_2 = -6$, and the leading coefficient $a$ is given as 1.

So, the factored form of the quadratic is: $f(x) = (x + 8)(x + 6)$

Now, expand the expression to get it in standard form: $f(x) = (x + 8)(x + 6) = x^2 + 6x + 8x + 48 = x^2 + 14x + 48$

Thus, the quadratic function is: $f(x) = x^2 + 14x + 48$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How do you find the zeros of a quadratic function?
2. What is the factored form of a quadratic equation?
3. Can a quadratic function have complex zeros?
4. How do you determine the vertex of a quadratic function?
5. What is the axis of symmetry of a quadratic function?

Tip: Always check the expanded form after factoring to ensure correctness.