What is the tangent of g(x)=x^2-8,\at x=-3 ?

The tangent of g(x)=x^2-8,\at x=-3 is y=-6x-17

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of g(x)=x^2-8,\at x=-3

tangent of

g (x) = x^{2} - 8, at x = - 3

y = - 6 x - 17

Solution steps

Find the tangent point:

(- 3, 1)

Find the slope of

g (x) = x^{2} - 8 : \frac{d g ( x )}{d x} = 2 x

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 = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x - 17

x^{2} y^{^{''}} + x y^{^{'}} + 4 y = 0

t an g e n t f (x) = x^{2} - 5 x + 5, (0, 5)

\frac{d}{d t} (e^{t} (cos (t)))

\int_{0}^{\frac{3}{4}} \frac{1}{9 + 16 x ^{2}} d x

\frac{d}{d x} (\frac{1}{2 π} e^{- \frac{x ^{2}}{2}})

What is the tangent of g(x)=x^2-8,\at x=-3 ?
The tangent of g(x)=x^2-8,\at x=-3 is y=-6x-17