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

The tangent of y=5x^2 is y=10a_{0}x-5a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=5x^2

tangent of

y = 5 x^{2}

y = 10 a_{0} x - 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 5 x^{2} : \frac{d y}{d x} = 10 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 10 a_{0}

Find the line with slope m=

10 a_{0}

and passing through

(a_{0}, 5 a_{0}^{2}) : y = 10 a_{0} x - 5 a_{0}^{2}

y = 10 a_{0} x - 5 a_{0}^{2}

\int_{- 1}^{1} - x d x

\int_{- 5}^{5} ∣ x ∣ d x

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

x^{^{''}} - 5 x = 0

2 5^{^{'}}