@Batterydude6000, I can tell you are taking an AoPS class, so I won't be sharing the answers but I'll give hints.
1 - We transform the equation into \(x = y^2 - 8y + 13\). We can then complete the square in \(y\), which turns the equation into \(x = (y-4)^2 - 3\). See if you can solve yourself from here.
2 - Again, we can complete the square to get \(y = -2(x^2-4x)-15\). This simplifies to \(y = -2(x - 2)^2 - 7\). Try solving it from here.
3 - To complete the square, we can add \(6y \) and subtract \(2x\), then add \(1 \) to complete the square for \(x\) and add \(9\) to complete the square for \(y\). It should be pretty simple from here.
4 - Since the axis of symmetry is \(x = 4\), we can substitute that into \(y = a(x-h)^2+k\).
5 - The simplified equation of \(y = a(x-h)^2+k\) is \(y = a(x + 3)^2 - 2\). We can also tell that the parabola passes through the point \((-1,0)\), so \(x = -1, ~ y = 0\). Try substituting that in the equation and sees where that gets you.
Hope that helped. 
AnxiousLlama
+26400 
