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

The tangent of y=(x^3+2)(3x^2-4x+2),(1,3) is y=9x-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = 9 x - 6

Solution steps

Find the slope of

y = (x^{3} + 2) (3 x^{2} - 4 x + 2) : \frac{d y}{d x} = 15 x^{4} - 16 x^{3} + 6 x^{2} + 12 x - 8

Compute the slope of

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

(1, 3) : m = 9

Find the line with slope m=

9

and passing through

(1, 3) : y = 9 x - 6

y = 9 x - 6

x^{2} (2.4)

\int 7 x + 5 d x

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

x \to 3 + lim (\frac{1}{( x - 3 ) ^{2}})

What is the tangent of y=(x^3+2)(3x^2-4x+2),(1,3) ?
The tangent of y=(x^3+2)(3x^2-4x+2),(1,3) is y=9x-6