What is the tangent of 4x^2-2x+3 ?

The tangent of 4x^2-2x+3 is y=(8a_{0}-2)x-4a_{0}^2+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 4x^2-2x+3

tangent of

4 x^{2} - 2 x + 3

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 2 a_{0} + 3)

Find the slope of

y = 4 x^{2} - 2 x + 3 : \frac{d y}{d x} = 8 x - 2

Compute the slope of

y = 4 x^{2} - 2 x + 3

x = a_{0} : m = 8 a_{0} - 2

Find the line with slope m=

8 a_{0} - 2

and passing through

(a_{0}, 4 a_{0}^{2} - 2 a_{0} + 3) : y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

\int_{0}^{π} sin (x) + cos (2 x) d x

y^{^{'}} + \frac{4}{5} y = 1 - \frac{1}{4} t, y (0) = 0

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

\frac{d}{d x} (- 1 - x)

\int_{0}^{\frac{1}{6}} x arcsin (6 x) d x

What is the tangent of 4x^2-2x+3 ?
The tangent of 4x^2-2x+3 is y=(8a_{0}-2)x-4a_{0}^2+3