What is the tangent of f(x)=x^2-2x,(5,9) ?

The tangent of f(x)=x^2-2x,(5,9) is y=2(4+sqrt(6))x-31-10sqrt(6) \lor y=2(4-sqrt(6))x+10sqrt(6)-31

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^2-2x,(5,9)

tangent of

f (x) = x^{2} - 2 x, (5, 9)

y = 2 (4 + 6) x - 31 - 106 or y = 2 (4 - 6) x + 106 - 31

Solution steps

Slope of the tangent line at

(5, 9) : m = \frac{x ^{2} - 2 x - 9}{x - 5}

Find the slope of

f (x) = x^{2} - 2 x : m = 2 x - 2

Solve

2 x - 2 = \frac{x ^{2} - 2 x - 9}{x - 5} : x = 5 + 6, x = 5 - 6

Find the tangent line at

x = 5 + 6 : y = 2 (4 + 6) x - 31 - 106

Find the tangent line at

x = 5 - 6 : y = 2 (4 - 6) x + 106 - 31

y = 2 (4 + 6) x - 31 - 106 or y = 2 (4 - 6) x + 106 - 31

x \to 0 lim (\frac{1 + x - 1}{x})

\int_{0}^{3} c \cdot (3 - x) d x

(1, - 2), (3, 4)

x \to 0 lim (x - 6)

\frac{d}{d x} (cos (x^{3} + 5 x))

What is the tangent of f(x)=x^2-2x,(5,9) ?
The tangent of f(x)=x^2-2x,(5,9) is y=2(4+sqrt(6))x-31-10sqrt(6) \lor y=2(4-sqrt(6))x+10sqrt(6)-31