Subtract these polynomials
2 answers:
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Answer:
(6, 9 ) and r = 3
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Given
x² + y² - 12x - 18y + 108 = 0
Rearrange the x- terms and the y- terms together and subtract 108 from both sides, that is
x² - 12x + y² - 18y = - 108
To obtain standard form use the method of completing the square
add ( half the coefficient of the x and y terms )² to both sides
x² + 2(- 6)x + 36 + y² + 2(- 9)y + 81 = - 108 + 36 + 81
(x - 6)² + (y - 9)² = 9 ← in standard form
with centre = (6, 9 ) and r = = 3
Answer: Choice B) (2,0)
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Set the first piece equal to zero and solve for x
x^2 - 4 = 0
x^2 = 4
x = sqrt(4) or x = -sqrt(4)
x = 2 or x = -2
Keep in mind that the first piece y = x^2-4 is only graphed when x = 2 or larger, so we ignore x = -2. This is one x intercept, but there may be more. Let's check the other graph to see what we get.
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Set the second piece equal to zero and solve for x
x-2 = 0
x-2+2 = 0+2
x = 2
We get the same result as above in the prior section. Because of this, the two pieces connect at this junction point.
The x intercept x = 2 leads to the location (2,0). The x intercept always occurs when y = 0.
The graph below shows this.
side note: the red piece y = x^2 - 4 looks linear, but it's actually not a straight line. It's just a really stretched out curve.
