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

 Nov 9, 2020
 #1
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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

 Nov 9, 2020
 #3
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I usually prefer using the vertex method, where all you have to do is rearrange the equation.

 

But this is nice! This person will now know two methods to do these problems! laugh

ETERNITY  Nov 9, 2020
 #2
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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! angel

 Nov 9, 2020
 #4
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I like this one better. It's faster and easier to do. Thank you!

Guest Nov 9, 2020
 #5
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No problem! Always happy to help :D

ETERNITY  Nov 9, 2020

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