What is the tangent of f(x)=-3x^2+2,\at x=1 ?

The tangent of f(x)=-3x^2+2,\at x=1 is y=-6x+5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=-3x^2+2,\at x=1

tangent of

f (x) = - 3 x^{2} + 2, at x = 1

y = - 6 x + 5

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

f (x) = - 3 x^{2} + 2 : \frac{df ( x )}{d x} = - 6 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

(1, - 1) : y = - 6 x + 5

y = - 6 x + 5

f (x) = \frac{4 x ^{3}}{5 - 2 x}

\frac{\partial}{\partial x} (e^{2 x yz - 7})

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

x \to 7 lim (sec (\frac{π x}{6}))

\frac{d y}{d x} = 7 x + y

What is the tangent of f(x)=-3x^2+2,\at x=1 ?
The tangent of f(x)=-3x^2+2,\at x=1 is y=-6x+5