Find the slope int. equation of the normal line to the graph f(x)=x^3-x^2 at the point where x=1?
- Marley KLv 74 weeks agoFavourite answer
first find the slope of the tangent line by taking the derivative:
f(x) = x^3 - x^2
f'(x) = 3x^2 - 2x
f'(1) = 3 - 2 = 1 <<<====slope of tangent line where x=1
the normal line is perpendicular to the tangent, so the slope of the normal line is -1.
find the y-coordinate:
f(1) = 1 - 1 = 0
Now you have the slope, m, and point (1,0) and can use that to find the equation:
y = mx + b
0 = (-1)(1) + b
b = 1
equation in slope-intercept form:
y = -1x + 1