What is the equation of the line normal to #f(x)=(4-2x)^2 # at #x=-3#?
1 Answer
Explanation:
First, find the point the normal line will intercept.
#f(-3)=(4+6)^2=100#
The normal line will pass through the point
To find the slope of the normal line, we must first know the slope of the tangent line. The slope of the tangent line can be found through calculating
First, find
#f'(x)=2(4-2x)*(-2)=-4(4-2x)#
The slope of the tangent line is
#f'(-3)=-4(4+6)=-40#
However, the normal line is perpendicular to the tangent line. Perpendicular lines have opposite reciprocal slopes. Thus, the slope of the normal line is the opposite reciprocal of
Relate the point the normal line intercepts,
#y-100=1/40(x+3)#