What is the equation of the line normal to $f(x)=(x-3)^2+4x$ at $x=1$ ?

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Answer 1 · 2018-04-19T05:29:18

Slope of normal is $x=1$

The slope of tangent at a point $x=x_0$ on $f(x)$ is $f'(x_0)=((df)/(dx))_(x=x_0)$

and as normal is perpendicular to slope, its slope is $-1/(f'(x_0))$

Here we are seeking normal at $x=1$ on curve $f(x)=(x-3)^2+4x$ i.e. at $(1,(1-3)^2+4*1)$ or $(1,8)$ .

As $(df)/(dx)=2(x-3)+4=2x-2$ , slope of tangent at $x=1$ is $0$ i.e. tangent at $(1,8)$ is $y=8$

and normal at $(1,8)$ would be $x=1$

What is the equation of the line normal to f(x)=(x-3)^2+4x f(x)=(x-3)^2+4x at x=1 x=1?