What is the tangent of f(x)=x^2+8,(-3,17) ?

The tangent of f(x)=x^2+8,(-3,17) is y=-6x-1

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^2+8,(-3,17)

tangent of

f (x) = x^{2} + 8, (- 3, 17)

y = - 6 x - 1

Solution steps

Find the slope of

f (x) = x^{2} + 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 8

(- 3, 17) : m = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x - 1

x \to - 2 lim (4 x)

y^{^{'}} - 2 (x + 1)^{3} = \frac{4}{( x + 1 )} y

\frac{d}{d x} (x^{5} x)

f (x) = 3 x - 4

\int_{- \infty}^{1} e^{4 x} d x

What is the tangent of f(x)=x^2+8,(-3,17) ?
The tangent of f(x)=x^2+8,(-3,17) is y=-6x-1