What is the normal of y=(-4)+x^2,(1,-3) ?

The normal of y=(-4)+x^2,(1,-3) is y=-1/2 x-5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

normal of y=(-4)+x^2,(1,-3)

normal of

y = (- 4) + x^{2}, (1, - 3)

y = - \frac{1}{2} x - \frac{5}{2}

Solution steps

Find the slope of

y = (- 4) + x^{2} : \frac{d y}{d x} = 2 x

Compute the slope of

y = (- 4) + x^{2}

(1, - 3) : m = 2

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(1, - 3) : y = - \frac{1}{2} x - \frac{5}{2}

y = - \frac{1}{2} x - \frac{5}{2}

\int x^{2} x + 3 d x

cos (2 x)

\frac{\partial}{\partial z} (x^{2} + 2 y + 2 z)

\frac{d}{d x} (\frac{7}{x ^{4}})

x \to 0 lim (- 2 csc (2 x))

What is the normal of y=(-4)+x^2,(1,-3) ?
The normal of y=(-4)+x^2,(1,-3) is y=-1/2 x-5/2