What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=-3 ?

The tangent of f(x)=x^3-6x^2-34x+40,\at x=-3 is y=29x+148

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-6x^2-34x+40,\at x=-3

tangent of

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = - 3

y = 29 x + 148

Solution steps

Find the tangent point:

(- 3, 61)

Find the slope of

f (x) = x^{3} - 6 x^{2} - 34 x + 40 : \frac{df ( x )}{d x} = 3 x^{2} - 12 x - 34

Compute the slope of

f (x) = x^{3} - 6 x^{2} - 34 x + 40

x = - 3 : m = 29

Find the line with slope m=

29

and passing through

(- 3, 61) : y = 29 x + 148

y = 29 x + 148

\int_{0}^{7} x 49 - x^{2} d x

\int_{0}^{2} x (x^{2} - 1)^{3} d x

\frac{\partial}{\partial x} (x^{2} y + 4 x^{4} y)

n = 3 \sum \infty \frac{1}{n ( n + 1 )}

\int \frac{a e ^{θ} + b}{a e ^{θ} - b} d θ

What is the tangent of f(x)=x^3-6x^2-34x+40,\at x=-3 ?
The tangent of f(x)=x^3-6x^2-34x+40,\at x=-3 is y=29x+148