Identify the vertex of the parabola xsquared-8x-28y-124=0
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To identify the vertex of the parabola given by the equation xsquared - 8x - 28y - 124 = 0, we need to rewrite the equation in vertex form, which is of the general form (x - h)^2 = 4a(y - k), where (h, k) represents the vertex.
Let's begin by rearranging the equation:
x^2 - 8x - 28y - 124 = 0
To complete the square, we need to group the x terms and the constant terms separately:
(x^2 - 8x) - 28y - 124 = 0
Next, we need to add and subtract a constant term to complete the square within the parentheses. In this case, since the coefficient of x is -8, we divide it by 2 and square the result (-8/2)^2 = 16:
(x^2 - 8x + 16 - 16) - 28y - 124 = 0
We can now combine the terms inside the parentheses:
(x - 4)^2 - 16 - 28y - 124 = 0
Simplifying further:
(x - 4)^2 - 28y - 140 = 0
Now, let's rearrange the equation to match the vertex form:
(x - 4)^2 = 28y + 140
Comparing this with the general form (x - h)^2 = 4a(y - k), we can determine that the vertex (h, k) is (4, -5). Therefore, the vertex of the given parabola is (4, -5).
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