Finding the X-coordinate of a normal to the tangent

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SUMMARY

The discussion centers on finding the X-coordinate of a point Q on curve C, defined by the equation y = x³ - 2x² - x + 9, which is perpendicular to the tangent line at point P(2,7) with the equation y = 3x + 1. The gradient of the normal line is confirmed to be -1/3, as it is the negative reciprocal of the tangent's slope of 3. The solution requires demonstrating that the X-coordinate of point Q is given by the expression [1/3(2+√6)]. The necessary derivative, dy/dx = 3x² - 4x - 1, is also provided for further calculations.

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Homework Statement



Tangent to C at point P(2,7) has an equation of y=3x+1.

Point Q also lies on C and is perpendicular to the tangent,
show that the X-coordinate is [1/3(2+√6)]

Homework Equations


curve C has equation y= x3-2x2-x+9
dy/dx = 3x2-4x-1

The Attempt at a Solution


gradient of normal has to = -1/3, right? considering that the tangent that is is perpendicular to is +3
I'm not sure as, the point that the normal is passing through hasn't been specified.

Here is the actual question: http://i259.photobucket.com/albums/hh299/the-real-guitar-hero/CapturePNGnnn_zpsd170ce09.png

Part C.
 
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Count Duckula said:

Homework Statement



Tangent to C at point P(2,7) has an equation of y=3x+1.

Point Q also lies on C and is perpendicular to the tangent,
show that the X-coordinate is [1/3(2+√6)]

Homework Equations


curve C has equation y= x3-2x2-x+9
dy/dx = 3x2-4x-1

The Attempt at a Solution


gradient of normal has to = -1/3



I'm not sure as the point, that the normal is passing through hasn't been specified.

Here is the actual question: http://i259.photobucket.com/albums/hh299/the-real-guitar-hero/CapturePNGnnn_zpsd170ce09.png

Part C.

Problems involving derivatives should be posted in the Calculus & Beyond section. I am moving this thread to that section.
 

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