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.

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

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

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