What is the tangent of Y(x)=x^2+3 ?

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

tangent of

Y (x) = x^{2} + 3

y = 2 a_{0} x - a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

Y (x) = x^{2} + 3 : \frac{d Y ( x )}{d x} = 2 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 = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

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

y = 2 a_{0} x - a_{0}^{2} + 3

\frac{t ^{2} + 7}{( 6 t - 2 ) ^{6}}

x \to 6 - lim (c x^{2} + 9 x)

\int sin (\frac{n x π}{2}) d x

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (e^{x})

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

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