Asked by Anonymous
Find the equation of the line(s) tangent to the curve y = x
3 − 6x + 2
and parallel to the line y = 6x − 2 .
3 − 6x + 2
and parallel to the line y = 6x − 2 .
Answers
Answered by
oobleck
I assume you meant y = x^3 - 6x + 2
At any point, the slope is y' = 3x^2 - 6
So you want the points where the slope is 6
That would be where
3x^2 - 6 = 6
x = ±2
So the points are (-2,6) and (2,-2)
That makes the lines
y-6 = 6(x+2)
y+2 = 6(x-2)
At any point, the slope is y' = 3x^2 - 6
So you want the points where the slope is 6
That would be where
3x^2 - 6 = 6
x = ±2
So the points are (-2,6) and (2,-2)
That makes the lines
y-6 = 6(x+2)
y+2 = 6(x-2)
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