What is the tangent of f(x)=2x^2+3x,(-3,9) ?

The tangent of f(x)=2x^2+3x,(-3,9) is y=-9x-18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2x^2+3x,(-3,9)

tangent of

f (x) = 2 x^{2} + 3 x, (- 3, 9)

y = - 9 x - 18

Solution steps

Find the slope of

f (x) = 2 x^{2} + 3 x : \frac{df ( x )}{d x} = 4 x + 3

Compute the slope of

f (x) = 2 x^{2} + 3 x

(- 3, 9) : m = - 9

Find the line with slope m=

- 9

and passing through

(- 3, 9) : y = - 9 x - 18

y = - 9 x - 18

\frac{d}{d x} ((x^{3} - 1)^{4})

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

\frac{19 x ^{2}}{2}

\frac{\partial}{\partial x} (2 x^{3} y^{2} cos (6 x^{3} y))

\int \frac{7}{sec ( x ) - 1} d x

What is the tangent of f(x)=2x^2+3x,(-3,9) ?
The tangent of f(x)=2x^2+3x,(-3,9) is y=-9x-18