What is the equation of the normal line of #f(x)=x^2 # at #x=5 #?

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Answer 1 · 2015-12-04T02:35:40

#y = -1/10x + 51/2#

The normal line is the line perpendicular to the tangent line. Then, the steps are as follows:

Find the slope #m_"tangent"# of tangent line
-Find the first derivative of the function
-Evaluate the first derivative at the desired point
Find the slope #m_"norm"# of the normal line
-Let #m_"norm" = -1/m_"tangent"#
Find the line with slope #m_"norm"# passing through the given point
-Using point-slope form: #(y-y_1) = m_"norm"(x-x_1)#

Let's go through the process.

#f'(x) = d/dxx^2 = 2x#

#m_"tangent" = f'(5) = 2(5) = 10#

#m_"norm" = -1/m_"tangent" = -1/10#

#(x_1, y_1) = (5, f(5)) = (5, 25)#

Substituting, we get the equation of the normal line.

#y - 25 = -1/10(x - 5)#

#=> y = -1/10x + 51/2#