What is the tangent of f(x)=2x^2-5x+3,\at a=1 ?

The tangent of f(x)=2x^2-5x+3,\at a=1 is y=(4a-5)x-2a^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,\at a=1

tangent of

f (x) = 2 x^{2} - 5 x + 3, at a = 1

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

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, 2 a^{2} - 5 a + 3)

Find the slope of

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

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 = 4 a - 5

Find the line with slope m=

4 a - 5

and passing through

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

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

\frac{d}{d x} ((y^{2} + x^{2}) e^{y^{2} - x^{2}})

\frac{d}{d x} (- 20 x^{3} + 12 x^{2} - 20)

\frac{d}{d x} (arctan (\frac{5 x}{4}))

\frac{d}{d x} (d y (d, x))

\int e^{6 θ} sin (7 θ) d θ

What is the tangent of f(x)=2x^2-5x+3,\at a=1 ?
The tangent of f(x)=2x^2-5x+3,\at a=1 is y=(4a-5)x-2a^2+3