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

The tangent of f(x)=-8x^2-7,\at x=1 is y=-16x+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)=-8x^2-7,\at x=1

tangent of

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

y = - 16 x + 1

Solution steps

Find the tangent point:

(1, - 15)

Find the slope of

f (x) = - 8 x^{2} - 7 : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

f (x) = - 8 x^{2} - 7

x = 1 : m = - 16

Find the line with slope m=

- 16

and passing through

(1, - 15) : y = - 16 x + 1

y = - 16 x + 1

\int ln (5 x + 8) d x

\frac{x ^{2} \cdot cos ( x )}{( 2 x + 1 ) ^{3}}

\int \frac{5 x + 2}{3 x 1 - 3 x} d x

x \to 0 lim (sin (π x))

\int \frac{( in x ) ^{2}}{x} d x

What is the tangent of f(x)=-8x^2-7,\at x=1 ?
The tangent of f(x)=-8x^2-7,\at x=1 is y=-16x+1