What is the tangent of y=(7x^2-6x+3)(1+2x),\at x=1 ?

The tangent of y=(7x^2-6x+3)(1+2x),\at x=1 is y=32x-20

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=(7x^2-6x+3)(1+2x),\at x=1

tangent of

y = (7 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

y = 32 x - 20

Solution steps

Find the tangent point:

(1, 12)

Find the slope of

y = (7 x^{2} - 6 x + 3) (1 + 2 x) : \frac{d y}{d x} = 42 x^{2} - 10 x

Compute the slope of

y = (7 x^{2} - 6 x + 3) (1 + 2 x)

x = 1 : m = 32

Find the line with slope m=

32

and passing through

(1, 12) : y = 32 x - 20

y = 32 x - 20

\int \frac{1}{4} d x

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

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

x^{2} y^{^{'}} = 3 x y + 4 x^{6}

h \to 0 lim (\frac{∣ 3 + h ∣ - ∣ 3 ∣}{h})

What is the tangent of y=(7x^2-6x+3)(1+2x),\at x=1 ?
The tangent of y=(7x^2-6x+3)(1+2x),\at x=1 is y=32x-20