What is the tangent of f(x)=2(x-3)^2+1,(4,3) ?

The tangent of f(x)=2(x-3)^2+1,(4,3) is y=4x-13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2(x-3)^2+1,(4,3)

tangent of

f (x) = 2 (x - 3)^{2} + 1, (4, 3)

y = 4 x - 13

Solution steps

Find the slope of

f (x) = 2 (x - 3)^{2} + 1 : \frac{df ( x )}{d x} = 4 x - 12

Compute the slope of

f (x) = 2 (x - 3)^{2} + 1

(4, 3) : m = 4

Find the line with slope m=

4

and passing through

(4, 3) : y = 4 x - 13

y = 4 x - 13

\frac{d}{d t} (cos (\frac{t ^{3}}{10}))

x \to 2 lim (\frac{9 x}{x - 2})

\int x^{2} \cdot 3^{2 x} d x

n = 1 \sum \infty \frac{1}{n ^{0.7}}

\int \frac{1}{- 7 - 8 x - x ^{2}} d x

What is the tangent of f(x)=2(x-3)^2+1,(4,3) ?
The tangent of f(x)=2(x-3)^2+1,(4,3) is y=4x-13