What is the tangent of y=x^2+5 ?

The tangent of y=x^2+5 is y=2a_{0}x-a_{0}^2+5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^2+5

tangent of

y = x^{2} + 5

y = 2 a_{0} x - a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{2} + 5 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 5

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

Find the line with slope m=

2 a_{0}

and passing through

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

y = 2 a_{0} x - a_{0}^{2} + 5

- cos (3 x) 3

\frac{d}{d x} (\frac{x ^{2} - 1}{4 - x ^{2}})

\int \frac{7}{x ( x ^{4} + 2 )} d x

(ln (x))^{x}

\int 4 t^{2} + 9 t^{4} d t