What is the tangent of f(x)=x^2+7,(1,8) ?

The tangent of f(x)=x^2+7,(1,8) is y=2x+6

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+7,(1,8)

tangent of

f (x) = x^{2} + 7, (1, 8)

y = 2 x + 6

Solution steps

Find the slope of

f (x) = x^{2} + 7 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 7

(1, 8) : m = 2

Find the line with slope m=

2

and passing through

(1, 8) : y = 2 x + 6

y = 2 x + 6

\int 2 sin (x) - 3 sec (x) tan (x) d x

\int \frac{t ^{5}}{( t ^{2} + 4 ) ^{2}} d t

\frac{d}{d x} (ln (arccos (x)))

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

\frac{d x}{d t} = t^{\frac{3}{2}}, x (3) = 7

What is the tangent of f(x)=x^2+7,(1,8) ?
The tangent of f(x)=x^2+7,(1,8) is y=2x+6