What is the equation of the line that is normal to #f(x)=-2x^2+4x-2 # at # x=3 #?
1 Answer
Explanation:
To find the normal line's slope, first find the slope of the tangent line at the same point. You can find the tangent line's slope by finding the value of the derivative at
The derivative of the function can be found through the power rule:
#f(x)=-2x^2+4x-2#
#f'(x)=-4x+4#
The slope of the tangent line is
#f'(3)=-4(3)+4=-8#
Now, to find the slope of the normal line, take the opposite reciprocal of
The opposite reciprocal of
The normal line intercepts the function at the point
(Don't be confused by the fact that both
The equation of the normal line can be written in point-slope form:
#y+8=1/8(x-3)#
In slope-intercept form, this is
#y=1/8x-67/8#
Graphed are the function and its normal line:
graph{(y+2x^2-4x+2)(y-x/8+67/8)((x-3)^2+(y+8)^2-.1)=0 [-15.11, 20.93, -15.42, 2.59]}