#1**0 **

x = y^2 -8y+13 is a sideways parabola

the vertex y coordinate is - b/2a where b = - 8 a = 1

use this value in the equation to calculate the x coordinate of the vertex

Guest Nov 9, 2020

#2**+1 **

All right... Here is my attempt:

We need to rewrite this equation in vertex form. First, we isolate the $x$ to give $x = y^2 - 8y + 13$. Completing the square in $y$, we get \[x = (y - 4)^2 - 3.\]

The graph of this equation is a rightward-opening parabola, so the vertex is the leftmost point, not the top or bottom point as usual.

Since $(y-4)^2$ is nonnegative, the value of $x$ is smallest when $y=4$, which makes $x=-3$. Therefore, the vertex of the parabola is $\boxed{(-3,4)}$.

Hope this helped!

ETERNITY Nov 9, 2020