Find the quadratic function, in both general and standard form, that has the vertex of $(-3,5)$ and passes through the point $(-6,-1)$.

Step 1 ：The general form of a quadratic function is \(y = ax^2 + bx + c\) and the standard form is \(y = a(x-h)^2 + k\), where \((h,k)\) is the vertex of the parabola.

Step 2 ：We know the vertex is \((-3,5)\), so we can substitute \(h=-3\) and \(k=5\) into the standard form to get \(y = a(x+3)^2 + 5\).

Step 3 ：We also know that the function passes through the point \((-6,-1)\). We can substitute \(x=-6\) and \(y=-1\) into the equation to solve for \(a\).

Step 4 ：Solving the equation \(-1 = 9a + 5\) gives us \(a = -\frac{2}{3}\).

Step 5 ：Now we can substitute \(a=-\frac{2}{3}\), \(h=-3\) and \(k=5\) into the standard form to get the equation of the parabola in standard form.

Step 6 ：We can also substitute \(a=-\frac{2}{3}\) into the general form to get the equation of the parabola in general form.

Step 7 ：Final Answer: The quadratic function that has the vertex of \((-3,5)\) and passes through the point \((-6,-1)\) is \(\boxed{y = -\frac{2}{3}(x+3)^2 + 5}\) in standard form and \(\boxed{y = -\frac{2}{3}x^2 - 4x - 1}\) in general form.