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# Vertex

0
3
1
+794

How to find vertex?

Find the vertex of the graph of the equation x - y^2 + 8y = 13 + 2y + y^2 + 7x - 11.

Dec 17, 2023

#1
+222
0

Sure, I can help you find the vertex of the graph of the equation.

First, we need to combine the like terms on both sides of the equation:

x + 7x - y^2 + y^2 = 13 + 11 + 2y + 8y

8x = 24 + 10y

Move the y term to the other side:

8x - 10y = 24

Now, we can rewrite the equation to isolate y:

y = (8x - 24) / 10

From this form, we see that the equation defines a parabola with its axis of symmetry parallel to the y-axis. To find the vertex, we need to rewrite it in vertex form:

y = a(x - h)^2 + k

where (h, k) is the vertex.

The coefficient of the squared term is not readily apparent, so we need to complete the square for the x term first:

y = (8/10)x - (24/10)

y = (4/5)x - (12/5)

Square both sides of the equation:

y^2 = (4/5)^2 x^2 - 2 * (4/5) * (12/5) x + (12/5)^2

Now, we can rewrite the expression as a squared term plus a constant:

y^2 = (4/5)^2 (x - 3)^2 - (144/25)

Move the constant term to the right side:

y^2 = (4/5)^2 (x - 3)^2 + (144/25 - 144/25)

y^2 = (4/5)^2 (x - 3)^2

Comparing this to the vertex form equation, we can see that:

h = 3

k = 0

Therefore, the vertex of the parabola is (3, 0).

Dec 17, 2023