What is the normal of y=sqrt(x),(4,2) ?

The normal of y=sqrt(x),(4,2) is y=-4x+18

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normal of y=sqrt(x),(4,2)

normal of

y = x, (4, 2)

y = - 4 x + 18

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{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 = \frac{1}{4}

Compute the slope of the perpendicular line:

m_{p} = - 4

Find the line with slope m=

- 4

and passing through

(4, 2) : y = - 4 x + 18

y = - 4 x + 18

x \to \infty lim (2^{- x})

\int 5 sin^{4} (x) d x

d er i v a t i v e y = x (2 x + 1)^{3}

\int 10 - t d t

d er i v a t i v e y = x^{- 2}

What is the normal of y=sqrt(x),(4,2) ?
The normal of y=sqrt(x),(4,2) is y=-4x+18