The problem asks you to write a quadratic function with zeros at -1 and 6, using the variable $x$ and in standard form, with a leading coefficient of 1.

To form the quadratic function from its zeros, we use the fact that if a quadratic function has roots (or zeros) at $r_1$ and $r_2$, the function can be written as:

$g(x) = a(x - r_1)(x - r_2)$

Given that the zeros are $-1$ and $6$, and the leading coefficient $a = 1$, the function becomes:

$g(x) = (x + 1)(x - 6)$

Now, expand this expression:

$g(x) = (x + 1)(x - 6) = x^2 - 6x + x - 6 = x^2 - 5x - 6$

Thus, the quadratic function is:

$g(x) = x^2 - 5x - 6$

Would you like more details, or do you have any questions?

Here are 5 questions for further practice:

1. What is the quadratic function with zeros at 3 and -2?
2. How would you factor the quadratic $x^2 + 2x - 8$?
3. What is the vertex of the function $g(x) = x^2 - 5x - 6$?
4. How do you find the axis of symmetry of a quadratic function?
5. How would you graph the function $g(x) = x^2 - 5x - 6$?

Tip: When you know the roots of a quadratic, you can always write it in factored form and expand to get the standard form.