What is the tangent of f(x)=3-8x^2,(-3,-69) ?

The tangent of f(x)=3-8x^2,(-3,-69) is y=48x+75

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=3-8x^2,(-3,-69)

tangent of

f (x) = 3 - 8 x^{2}, (- 3, - 69)

y = 48 x + 75

Solution steps

Find the slope of

f (x) = 3 - 8 x^{2} : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

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

(- 3, - 69) : m = 48

Find the line with slope m=

48

and passing through

(- 3, - 69) : y = 48 x + 75

y = 48 x + 75

\frac{d}{d x} (arctan (3 x^{3}))

\frac{x - 1}{x + 1}

\int \frac{cos ( x )}{2 sin ( x ) + 1} d x

\int x \cdot (4 - x) d x

5 + 4 δ (t)

What is the tangent of f(x)=3-8x^2,(-3,-69) ?
The tangent of f(x)=3-8x^2,(-3,-69) is y=48x+75