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

 Feb 12, 2024
 #1
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x - y^2 + 8y = 13 + 2y + y^2 - 7y - 14, combining like terms:

x = 2y^2 - 13y - 1

 

Note that this is a function of y, so f(y) = x, which will look like a sideways parabola when graphed. Still, the vertex can be calculated in the same manner as if it was f(x) = y.

Vertex form can be written as a(x - h)^2 + k = 0, where (h, k) is the vertex. In this problem specifically, the x = y, and the variables can be solved for using "complete the square". However, note that for our problem specifically, the vertex would be (k, h) because the this parabola is the inverse of a(x - h)^2 + k = 0.

 

2(y - 13/4)^2 - 169/8 - 1 = 2(y - 13/4)^2 - 177/8 = 0 

(h, k) = (13/4, -177/8), so the vertex to our problem, (k, h) = (-177/8, 13/4)

 Feb 12, 2024

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