What is the tangent of e^x(7-5x+3x^2),\at x=0 ?

The tangent of e^x(7-5x+3x^2),\at x=0 is y=2x+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of e^x(7-5x+3x^2),\at x=0

tangent of

e^{x} (7 - 5 x + 3 x^{2}), at x = 0

y = 2 x + 7

Solution steps

Find the tangent point:

(0, 7)

Find the slope of

y = e^{x} (7 - 5 x + 3 x^{2}) : \frac{d y}{d x} = 3 e^{x} x^{2} + e^{x} x + 2 e^{x}

Compute the slope of

y = e^{x} (7 - 5 x + 3 x^{2})

x = 0 : m = 2

Find the line with slope m=

2

and passing through

(0, 7) : y = 2 x + 7

y = 2 x + 7

\int 6 sec^{6} (x) d x

h (x) = (\frac{2}{x} - 3) (x^{2} + 7)

x \to 0 lim (ln (\frac{1 + x}{x}))

\int (x^{2} cos (x)) d x

\int sin (\frac{n \cdot π}{3} x) d x

What is the tangent of e^x(7-5x+3x^2),\at x=0 ?
The tangent of e^x(7-5x+3x^2),\at x=0 is y=2x+7