+27 Find the vertex of the graph of the equation $x - y^2 + 8y = 13.$
Find the vertex of the graph of the equation $y = -2x^2 + 8x - 15.$
Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$
The graph of $y = ax^2 + bx + c$ has an axis of symmetry of $x = 4.$ Find $\frac{b}{a}$.
The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$. (The distance between the grid lines is one unit.) 
+102 @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
+37153 1.) x = y^2 -8y + 13
re-arrange to vertex form
x = (y-4)^2 -3 vertex is -3,4
+37153 2 ) -2x^2+8x-15 re-arrange to vertex form
-2 (x^2 - 4x) -15
-2 ( x-2)^2 - 7 vertex = 2, -7
+37153 3) x^2 -2x + y^2 + 6y = 6 Arrange into standard circle equation form
(x-1)^2 + ( y+3)^2 = 6 +1 + 9
6+ 1 + 9 = 16 = r^2
Area of a circle = pi r^2 = pi (16) = 16 pi units2
+27 Can you answer the last two questions? Thank you so much for the other answers.... all correct!
