SOLUTION: How do you find the tangent and normal lines of something that starts out as x^2-4x and has a derivative of 2x-4
please help in the solving of this equation.
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Question 237520: How do you find the tangent and normal lines of something that starts out as x^2-4x and has a derivative of 2x-4
please help in the solving of this equation.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You have to have a place where you want the tangent and normal lines defined. That is, you can specifically answer the question: "Derive equations of the tangent and normal lines to the graph of \ =\ x^2\ -\ 4x)
at the point \right))
"
The first derivative is, as you stated, 
, and tells us the slope of the tangent line at any value of 
in the domain of the original function.
So, in general, at the point , the slope of the tangent line is 
. The normal line at the same point is the line perpendicular to the tangent at that point. Since perpendicular lines have negative reciprocals, that is:
We can say that the slope of the normal line at is 
Now all you need is the point-slope form of the equation of a straight line to derive the desired equations:
)
 = (2a\ -\ 4)(x - a)\ \ \ \ )
: Tangent
 = \frac{a\ -\ x}{2a\ -\ 4}\ \ \ \ )
: Normal
John
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