What is the normal of y=(3x+1)^2,(0,1) ?

The normal of y=(3x+1)^2,(0,1) is y=-1/6 x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

normal of y=(3x+1)^2,(0,1)

normal of

y = (3 x + 1)^{2}, (0, 1)

y = - \frac{1}{6} x + 1

Solution steps

Find the slope of

y = (3 x + 1)^{2} : \frac{d y}{d x} = 6 (3 x + 1)

Compute the slope of

y = (3 x + 1)^{2}

(0, 1) : m = 6

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(0, 1) : y = - \frac{1}{6} x + 1

y = - \frac{1}{6} x + 1

f (x) = x^{5}

f (x) = \frac{1}{x ^{4}}, (- 2, \frac{1}{16})

\frac{d}{d x} (\frac{c ^{2} - x ^{2}}{2 x})

\int π (\frac{1}{1 + x ^{2}})^{2} d x

\int_{0}^{1} \frac{2 t}{1 + t ^{2}} d t

What is the normal of y=(3x+1)^2,(0,1) ?
The normal of y=(3x+1)^2,(0,1) is y=-1/6 x+1