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 of

f (x) = x^{2} + 2, at x = 2

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

Compute the slope of

f (x) = x^{2} + 2

x = 2 : 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}

y = 6^{x}, at x = 1

f (x) = \frac{x}{1 + x ^{2}}

c = π^{8}

f (x) = x^{2}

(- 3, - 7), (3, - 5)

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