What is the equation of the line normal to $f(x)=4x^2-7x+6$ at $x=1$ ?

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Answer 1 · 2016-03-31T09:56:07

Equation of normal to $f(x)$ at $x=1$ is $x+y=4$ .

At $x=1$ , $f(x)=4-7+6=3$ , hence normal will be at point $(1,3)$ on the curve $f(x)=4x^2-7x+6$ .

The slope of tangent will be given by value of $f'(x)$ at $x=1$ /

As $f'(x)=8x-7$ , slope of tangent is $1$ .

As normal is perpendicular to tangent, slope of normal will be $-1/1=-1$ ;

Hence equation of normal given by point slope form will be

$(y-3)=-1*(x-1)$ or $y-3=-x+1$ or $x+y=4$ .

What is the equation of the line normal to f(x)=4x^2-7x+6 f(x)=4x^2-7x+6 at x=1x=1?