Home > Calculus > Normal Line to a Tangent

What is the equation of the normal line of #f(x)=x^2-4x# at #x=1#?

Answer 1

#y = 1/2x - 7/2#

A normal line is a perpendicular line. In order to find it at a certain point, we need to find the tangent line at that point, and then find the line passing through that point whose slope, multiplied by the slope of the tangent line, is equal to #-1#.

The slope of the tangent line at a point is given by the derivative of the function, evaluated at that point.

#f'(x) = d/dxx^2-4x = 2x-4# #=> f'(1) = 2(1) - 4 = -2#

Thus the slope of the tangent line of #f(x)# at #x=1# is #-2# and the slope of the normal line #m_n# has the property #-2m_n = -1# #=>m_n = 1/2#

The equation of a line with slope #m# passing through the point #(x_1, y_1)# is given by #y-y_1 = m(x-x_1)#. As we are looking for the normal line at #x=1#, we need it to pass through the point #(1, f(1)) = (1, -3)#.

Putting this together with the slope gives us #y + 3 = 1/2(x - 1)#

Thus the equation of the normal line is #y = 1/2x - 7/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Arocha

The equation of the normal line of f(x)=x^2-4x at x=1 is y = -3x + 1.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.