What is the equation of the line normal to $f(x)=(x-1)^2-2x+5$ at $x=-2$ ?

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Answer 1 · 2018-07-18T15:19:06

$x-8y+146=0$

The slope $f'(x)$ of tangent to the curve: $f(x)=(x-1)^2-2x+5$ is given by differentiating given function w.r.t. $x$ as follows

$f'(x)=d/dx((x-1)^2-2x+5)$

$=2(x-1)-2$

$=2x-4$

hence the slope of tangent at $x=-2$ is

$f'(-2)=2(-2)-4$

$=-8$

hence the slope of normal to the curve at the same point $x=-2$ is

$=-1/{-8}$

$=1/8$

Now, substituting $x=-2$ in the given function: $f(x)=(x-1)^2-2x+5$ we get $y$ coordinate of the point as follows

$f(-2)=(-2-1)^2-2(-2)+5$

$=18$

Hence, the equation of normal to the curve at $(-2, 18)\equiv(x_1, y_1)$ & having a slope $m=1/8$ is given as

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

$y-18=1/8(x-(-2))$

$8y-144=x+2$

$x-8y+146=0$

What is the equation of the line normal to f(x)=(x-1)^2-2x+5f(x)=(x-1)^2-2x+5 at x=-2x=-2?