What is the tangent of f(x)=2x^2-10x+9,(2,-3) ?

The tangent of f(x)=2x^2-10x+9,(2,-3) is y=-2x+1

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^2-10x+9,(2,-3)

tangent of

f (x) = 2 x^{2} - 10 x + 9, (2, - 3)

y = - 2 x + 1

Solution steps

Find the slope of

f (x) = 2 x^{2} - 10 x + 9 : \frac{df ( x )}{d x} = 4 x - 10

Compute the slope of

f (x) = 2 x^{2} - 10 x + 9

(2, - 3) : m = - 2

Find the line with slope m=

- 2

and passing through

(2, - 3) : y = - 2 x + 1

y = - 2 x + 1

x \to 1 lim (\frac{( 1 )}{( 1 + x )} - \frac{( 3 )}{( 1 - x ^{(3)} )})

(x^{2} + y^{2}) d x + 2 x y d y = 0

\int 30 (16 - x^{2}) d x

\int \frac{x ^{3} - 5 x - 11}{x ^{2} - x - 6} d x

n = 1 \sum \infty \frac{5 n}{n ^{2} + 2}

What is the tangent of f(x)=2x^2-10x+9,(2,-3) ?
The tangent of f(x)=2x^2-10x+9,(2,-3) is y=-2x+1