Finding Parallel Tangent Line Points on Curve: y=1/3x^3-4x^2+18x+22

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SUMMARY

The problem involves finding two points on the curve defined by the equation y = (1/3)x^3 - 4x^2 + 18x + 22 where the tangent lines are parallel to the line represented by the equation 6x - 2y = 1. The slope of the given line is determined to be 3. To solve for the points, one must derive the slope of the tangent lines to the cubic curve and set this equal to 3, leading to a quadratic equation that can be solved for the x-coordinates of the points of tangency.

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Hey... I am kinda got stuck with this problem... can you guys give me a hint or something...

Problem says:

The tangentline to the curve y=1/3 x^3 - 4x^2 + 18x + 22 is parallel to the line 6x - 2y = 1 at 2 points on the curve... find the 2 points.

Well I was thinking about writing the function for another line which has the same lope with the one given... so it should be something like y = 3x + m

Now what?

Thanks for your help.
 
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You don't need to write an equation for another line, you just have to know what the slope of 6x - 2y = 1 is. Then you need to find an equation for the slopes of the lines tangent to your cubic curve. Then you have to solve (a quadratic) for when the slopes of the tangents equal the slope of 6x - 2y = 1.
 

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