What is the normal of y=sqrt(2x+1),\at x=4 ?

The normal of y=sqrt(2x+1),\at x=4 is y=-3x+15

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normal of y=sqrt(2x+1),\at x=4

normal of

y = 2 x + 1, at x = 4

y = - 3 x + 15

Solution steps

Find the tangent point:

(4, 3)

Find the slope of

y = 2 x + 1 : \frac{d y}{d x} = \frac{1}{2 x + 1}

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 = \frac{1}{3}

Compute the slope of the perpendicular line:

m_{p} = - 3

Find the line with slope m=

- 3

and passing through

(4, 3) : y = - 3 x + 15

y = - 3 x + 15

What is the normal of y=sqrt(2x+1),\at x=4 ?
The normal of y=sqrt(2x+1),\at x=4 is y=-3x+15