What is the tangent of y=(x-2)(x^2+4),(1,-5) ?

The tangent of y=(x-2)(x^2+4),(1,-5) is y=3x-8

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)(x^2+4),(1,-5)

tangent of

y = (x - 2) (x^{2} + 4), (1, - 5)

y = 3 x - 8

Solution steps

Find the slope of

y = (x - 2) (x^{2} + 4) : \frac{d y}{d x} = 3 x^{2} - 4 x + 4

Compute the slope of

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

(1, - 5) : m = 3

Find the line with slope m=

3

and passing through

(1, - 5) : y = 3 x - 8

y = 3 x - 8

ln (2)

\frac{d y}{d x} = \frac{y}{x + 1} + 4 x^{2} + 4 x

\int_{\frac{4}{3}}^{4} \frac{1}{1 - x ^{2}} d x

\int (x^{3} + 1)^{2} (3 x) d x

y = \frac{x}{x + 9}, (1, 0.1)

What is the tangent of y=(x-2)(x^2+4),(1,-5) ?
The tangent of y=(x-2)(x^2+4),(1,-5) is y=3x-8