Question 978149: the equation for the tangent line at point(1,0) in the curve y=3x²-4x+1 is, y= ? and the area surrounded by the curve the tangen line ans the y-axis is?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! the equation for the tangent line at point(1,0) in the curve y=3x²-4x+1 is, y= ?
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y' = 6x - 4 = the slope, m
m = 2
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y = 2(x - 1)
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and the area surrounded by the curve the tangen line ans the y-axis is?
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Find the x-ints of the parabola, the zeroes:
3x^2 - 4x + 1 = 0
(3x-1)*(x-1) = 0
x = 1/3, x = 1
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The bounded area is the triangle below the x-axis.
Its area = 1 sq unit.
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Add the area under the parabola from x = 0 to x = 1/3 is:
f(x) = 3x^2 - 4x + 1
INT = x^3 - 2x^2 + x (ignore the constant)
Area = INT(1/3) - INT(0)
= 1/27 - 2/9 + 1/3 - 0
= 4/27 sq units
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The area from 1/3 to 1 has to be subtracted:
Area = INT(1) - INT(1/3)
= [1/27 - 2/9 + 1/3] - (1 - 2 + 1)
= 4/27
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Total area = 1 sq unit
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I think the other tutor got the area for the parabola, the tangent and the x-axis. Not the y-axis.
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