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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

Compute the slope of

y = 2 x + 1

x = 4 : 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

(x - y - 1)^{3} = x

n = 1 \sum \infty \frac{x ^{n}}{3 ^{n}}

\frac{d}{d x} (\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9})

x \to 1 lim (\frac{x + 1}{x + 2})

f^{^{'}} (x) = x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1

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