What is the tangent of f(x)=x^2+6,(5,31) ?

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

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,(5,31)

tangent of

f (x) = x^{2} + 6, (5, 31)

y = 10 x - 19

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x - 19

\int e^{- t^{2}} d t

\int \frac{1}{z ^{2} a ^{2} - x ^{2}} d z

x^{3} \frac{16}{3}

f (x) = 2 sin (x) + sin^{2} (x)

6

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