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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

tangent of

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

y = (6 a + 2) x - 3 a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, 3 a^{2} + 2 a)

Find the slope of

y = (3 x^{2} + 2 x) : \frac{d y}{d x} = 6 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 = 6 a + 2

Find the line with slope m=

6 a + 2

and passing through

(a, 3 a^{2} + 2 a) : y = (6 a + 2) x - 3 a^{2}

y = (6 a + 2) x - 3 a^{2}

a re a y = x, y = x - 2, [0, 4]

\frac{\partial}{\partial x} (2 x^{2} + 4 y^{2} - 3 z^{2} - 11)

\int (cos (t)) d t

\int \frac{x ^{4} + 12 x ^{2} + 1}{x ^{5} + 20 x ^{3} + 5 x} d x

d er i v a t i v e - 9 cos (3 x)

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