What is the normal of f(x)=(x+2)sqrt(2x-3),\at x=2 ?

The normal of f(x)=(x+2)sqrt(2x-3),\at x=2 is f(x)=-1/5 x+22/5

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normal of f(x)=(x+2)sqrt(2x-3),\at x=2

normal of

f (x) = (x + 2) 2 x - 3, at x = 2

f (x) = - \frac{1}{5} x + \frac{22}{5}

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

f (x) = (x + 2) 2 x - 3 : \frac{df ( x )}{d x} = \frac{3 x - 1}{2 x - 3}

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 = 5

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5}

Find the line with slope m=

- \frac{1}{5}

and passing through

(2, 4) : f (x) = - \frac{1}{5} x + \frac{22}{5}

f (x) = - \frac{1}{5} x + \frac{22}{5}

What is the normal of f(x)=(x+2)sqrt(2x-3),\at x=2 ?
The normal of f(x)=(x+2)sqrt(2x-3),\at x=2 is f(x)=-1/5 x+22/5