What is the equation of the normal line of $f(x)=(x-9)^2/(x-1)$ at $x=-2$ ?

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Answer 1 · 2016-06-06T09:15:03

The normal line

$y=(9x)/55-6601/165$

The given curve is

$y=(x-9)^2/(x-1)$
at the point $x=-2$

Let us solve for the point $(x_1, y_1)$ first

Let $x_1=-2$

$y_1=(x_1-9)^2/(x_1-1)=(-2-9)^2/(-2-1)=121/-3=-121/3$

Our point $(x_1, y_1)=(-2, -121/3)$

Solve for the slope of the normal line $m_n=-1/m$

$m=d/dx((x-9)^2/(x-1))=((x-1)*2(x-9)-(x-9)^2*1)/(x-1)^2$

substitute $x=-2$

$m=((-2-1)*2(-2-9)-(-2-9)^2*1)/(-2-1)^2=-55/9$

$m_n=-1/m$

$m_n=-1/(-55/9)$

$m_n=9/55$

Let us solve for the normal line using $m_n$ and $(x_1, y_1)$

$y-y_1=m_n(x-x_1)$

$y-(-121/3)=9/55(x-(-2))$

$y=(9x)/55-6601/165$

Kindly see the graphs of $y=(x-9)^2/(x-1)$ and the normal line $y=(9x)/55-6601/165$

Desmos

God bless....I hope the explanation is useful.

What is the equation of the normal line of f(x)=(x-9)^2/(x-1)f(x)=(x-9)^2/(x-1) at x=-2x=-2?