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.