What is the tangent of f(x)=x^2-6,(5,19) ?

The tangent of f(x)=x^2-6,(5,19) is y=10x-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^2-6,(5,19)

tangent of

f (x) = x^{2} - 6, (5, 19)

y = 10 x - 31

Solution steps

Find the slope of

f (x) = x^{2} - 6 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 6

(5, 19) : m = 10

Find the line with slope m=

10

and passing through

(5, 19) : y = 10 x - 31

y = 10 x - 31

\frac{d}{d x} (x^{3} e^{2 x})

\frac{\partial}{\partial x} (4 y + 9 x^{2})

\frac{d y}{d x} = 17 x + y - 17

\frac{d}{d x} (x^{(\frac{1}{2})})

\frac{d}{d x} (\frac{a}{x ^{2}})

What is the tangent of f(x)=x^2-6,(5,19) ?
The tangent of f(x)=x^2-6,(5,19) is y=10x-31