What is the normal of f(x)=x^2+2,\at x=2 ?

The normal of f(x)=x^2+2,\at x=2 is f(x)=-1/4 x+13/2

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normal of f(x)=x^2+2,\at x=2

normal

f (x) = - \frac{1}{4} x + \frac{13}{2}

Solution steps

Find the tangent point:

(2, 6)

Find the slope of

f (x) = x^{2} + 2 : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(2, 6) : f (x) = - \frac{1}{4} x + \frac{13}{2}

f (x) = - \frac{1}{4} x + \frac{13}{2}

What is the normal of f(x)=x^2+2,\at x=2 ?
The normal of f(x)=x^2+2,\at x=2 is f(x)=-1/4 x+13/2