Solution: This question has a bit of a deliberate visual challenge to it--the -10 point shown is not the vertex!! You can verify this algebraically because the \(x\) coordinate should be precisely in the middle of the \(x\) intercepts, but \(0\neq \frac{5-6}{{2}}\)!

In any problem where you know the horizontal intercepts, the simplest choice of quadratic form is the factored form of the parabola. I get started with the intercepts and the parameter \(a\) below:

\[ y = a(x-6)(x+5) \]

Pay close attention to the factors vs. the zeros. A zero at \(x=-6\) results in the factor \((x+6)\)!

Now that we have the basic form, we can solve for the parameter \(a\) by using the given vertical intercept, \((0,-10)\) and substituting into our equation:

\[ \solve{ -10 &=&a(0-6)(0+5)\\ -10 &=&-30a\\ \dfrac{{1}}{{3}}&=&a } \]

Now that we have our missing parameter we can write down our final answer: \(y=\dfrac{{1}}{{3}}(x-6)(x+5)\).