Find the vertex of the graph of the equation $x - y^2 + 8y = 13 + 2y + y^2 - 7y - 14$.

tomtom Dec 9, 2023

#1**-1 **

Let's start by rewriting the equation to be in terms of y : $$ y^2 - 8y + x = 13 - 2y - x $$

Completing the square on the y terms, we get

$$ (y-4)^2 - 8y + x = 13 - 2y - x $$

$$(y-4)^2 = 13 + 2(y-4) + 2x $$

$$(y-4)^2 - 2(y-4) = 2x + 13 $$

(y−4−1)2=2x+13+1

(y−5)2=2x+14

We can see that y=5 is the axis of symmetry. Since the equation is in the form (y−h)2=k(x−h), the vertex is located at (h,h). Therefore, the vertex is at (5,5).

BuiIderBoi Dec 9, 2023