What is the tangent of 2x^3+8x^2-5x+9 ?

The tangent of 2x^3+8x^2-5x+9 is y=(6a_{0}^2+16a_{0}-5)x-4a_{0}^3-8a_{0}^2+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 2x^3+8x^2-5x+9

tangent of

2 x^{3} + 8 x^{2} - 5 x + 9

y = (6 a_{0}^{2} + 16 a_{0} - 5) x - 4 a_{0}^{3} - 8 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 8 a_{0}^{2} - 5 a_{0} + 9)

Find the slope of

y = 2 x^{3} + 8 x^{2} - 5 x + 9 : \frac{d y}{d x} = 6 x^{2} + 16 x - 5

Compute the slope of

y = 2 x^{3} + 8 x^{2} - 5 x + 9

x = a_{0} : m = 6 a_{0}^{2} + 16 a_{0} - 5

Find the line with slope m=

6 a_{0}^{2} + 16 a_{0} - 5

and passing through

(a_{0}, 2 a_{0}^{3} + 8 a_{0}^{2} - 5 a_{0} + 9) : y = (6 a_{0}^{2} + 16 a_{0} - 5) x - 4 a_{0}^{3} - 8 a_{0}^{2} + 9

y = (6 a_{0}^{2} + 16 a_{0} - 5) x - 4 a_{0}^{3} - 8 a_{0}^{2} + 9

8 - 2 x, x^{3} - 7 x^{2} + 12 x, [1, 4]

h \to 0 lim (h)

\int_{1}^{3} 9 r^{3} ln (r) d r

(sec^{3} (x))^{^{'}}

\int sin (- 7 x) d x

What is the tangent of 2x^3+8x^2-5x+9 ?
The tangent of 2x^3+8x^2-5x+9 is y=(6a_{0}^2+16a_{0}-5)x-4a_{0}^3-8a_{0}^2+9