What is the tangent of f(x)=sqrt(100-x^2),\at x=-6 ?

The tangent of f(x)=sqrt(100-x^2),\at x=-6 is y= 3/4 x+25/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(100-x^2),\at x=-6

tangent of

f (x) = 100 - x^{2}, at x = - 6

y = \frac{3}{4} x + \frac{25}{2}

Solution steps

Find the tangent point:

(- 6, 8)

Find the slope of

f (x) = 100 - x^{2} : \frac{df ( x )}{d x} = - \frac{x}{100 - x ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(- 6, 8) : y = \frac{3}{4} x + \frac{25}{2}

y = \frac{3}{4} x + \frac{25}{2}

t sin (1.74 t)

\int - \frac{1}{3 ( s - 1 ) ^{3}} d s

\int \frac{x}{x - 13} d x

\int cos^{4} (3 x) d x

y = \frac{d y}{d x} x - \frac{d y}{d x}^{2}

What is the tangent of f(x)=sqrt(100-x^2),\at x=-6 ?
The tangent of f(x)=sqrt(100-x^2),\at x=-6 is y= 3/4 x+25/2