What is the tangent of f(x)=x^2-4,\at x=-5,21 ?

The tangent of f(x)=x^2-4,\at x=-5,21 is y=-10x-29

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^2-4,\at x=-5,21

tangent of

f (x) = x^{2} - 4, at x = - 5, 21

y = - 10 x - 29

Solution steps

Find the tangent point:

(- 5, 21)

Find the slope of

f (x) = x^{2} - 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 4

x = - 5 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 21) : y = - 10 x - 29

y = - 10 x - 29

\int \frac{x}{3 x ^{2} + 1} d x

cos (x), \frac{π}{4}

f (x) = 2 sec (x) - 3 x

\int sin^{2} (x co) s^{4} x d x

x \to 0 + lim (sin (x) ln (3 x))

What is the tangent of f(x)=x^2-4,\at x=-5,21 ?
The tangent of f(x)=x^2-4,\at x=-5,21 is y=-10x-29