+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=y2−8y+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(x2−4x)−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
+37165 1.) x = y^2 -8y + 13
re-arrange to vertex form
x = (y-4)^2 -3 vertex is -3,4
+37165 2 ) -2x^2+8x-15 re-arrange to vertex form
-2 (x^2 - 4x) -15
-2 ( x-2)^2 - 7 vertex = 2, -7
+37165 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!
