What is the tangent of y=x^2,(-0.7,0.49) ?

The tangent of y=x^2,(-0.7,0.49) is y=-1.4x-0.49

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^2,(-0.7,0.49)

tangent of

y = x^{2}, (- 0.7, 0.49)

y = - 1.4 x - 0.49

Solution steps

Find the slope of

y = x^{2} : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2}

(- 0.7, 0.49) : m = - 1.4

Find the line with slope m=

- 1.4

and passing through

(- 0.7, 0.49) : y = - 1.4 x - 0.49

y = - 1.4 x - 0.49

\int (x - 8) sin (π x) d x

t^{2} - e^{- 9 t} + 5

\frac{d}{d x} (\frac{1}{2 x} - \frac{1}{3 x ^{2}})

\int sin (x) sec^{3} (x) d x

x y^{2} + 3 y^{2} - x^{2} \frac{d y}{d x} = 0

What is the tangent of y=x^2,(-0.7,0.49) ?
The tangent of y=x^2,(-0.7,0.49) is y=-1.4x-0.49