What is the tangent of y=-6x^2+5x,(0,0) ?

The tangent of y=-6x^2+5x,(0,0) is y=5x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=-6x^2+5x,(0,0)

tangent of

y = - 6 x^{2} + 5 x, (0, 0)

y = 5 x

Solution steps

Find the slope of

y = - 6 x^{2} + 5 x : \frac{d y}{d x} = - 12 x + 5

Compute the slope of

y = - 6 x^{2} + 5 x

(0, 0) : m = 5

Find the line with slope m=

5

and passing through

(0, 0) : y = 5 x

y = 5 x

\frac{d y}{d t} + y = cos (t), y (0) = 1

\frac{1}{2} x

\frac{d}{d x} (\frac{18 x + 2 x ^{2}}{( 9 + 2 x ) ^{2}})

\int \frac{1}{( 49 + x ^{2} ) ^{\frac{3}{2}}} d x

\int_{- 1}^{3} (x^{2} - 4 x + 3) d x

What is the tangent of y=-6x^2+5x,(0,0) ?
The tangent of y=-6x^2+5x,(0,0) is y=5x