What is the tangent of f(x)=x^2-9x+17,\at x=4 ?

The tangent of f(x)=x^2-9x+17,\at x=4 is y=-x+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)=x^2-9x+17,\at x=4

tangent of

f (x) = x^{2} - 9 x + 17, at x = 4

y = - x + 1

Solution steps

Find the tangent point:

(4, - 3)

Find the slope of

f (x) = x^{2} - 9 x + 17 : \frac{df ( x )}{d x} = 2 x - 9

Compute the slope of

f (x) = x^{2} - 9 x + 17

x = 4 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 1

x^{3} - 3 x + 3

\frac{8 x}{x ^{2} + 3}

n = 0 \sum \infty (\frac{1}{7})^{n}

\frac{\partial}{\partial x} (\frac{3 x}{5 y})

(6 x^{2} y^{2}) d x + (4 x^{3} y) d y = 0

What is the tangent of f(x)=x^2-9x+17,\at x=4 ?
The tangent of f(x)=x^2-9x+17,\at x=4 is y=-x+1