What is the tangent of f(x)=2x^2+5x-3 ?

The tangent of f(x)=2x^2+5x-3 is y=(4a_{0}+5)x-2a_{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 f(x)=2x^2+5x-3

tangent of

f (x) = 2 x^{2} + 5 x - 3

y = (4 a_{0} + 5) x - 2 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 5 a_{0} - 3)

Find the slope of

f (x) = 2 x^{2} + 5 x - 3 : \frac{df ( x )}{d x} = 4 x + 5

Compute the slope of

f (x) = 2 x^{2} + 5 x - 3

x = a_{0} : m = 4 a_{0} + 5

Find the line with slope m=

4 a_{0} + 5

and passing through

(a_{0}, 2 a_{0}^{2} + 5 a_{0} - 3) : y = (4 a_{0} + 5) x - 2 a_{0}^{2} - 3

y = (4 a_{0} + 5) x - 2 a_{0}^{2} - 3

\int \frac{1}{( x ^{2} - 1 ) ( x - 3 )} d x

\int 2 x - 2 d x

y^{^{'}} + \frac{y}{t} = t^{2} y^{4}

x^{2} + 5 x, at x = 4

x \to + 4 + lim (\frac{7}{x - 4})

What is the tangent of f(x)=2x^2+5x-3 ?
The tangent of f(x)=2x^2+5x-3 is y=(4a_{0}+5)x-2a_{0}^2-3