What is the tangent of y=3x^2+1,(0,-1) ?

The tangent of y=3x^2+1,(0,-1) is y=2sqrt(6)x-1 \lor y=-2sqrt(6)x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=3x^2+1,(0,-1)

tangent of

y = 3 x^{2} + 1, (0, - 1)

y = 26 x - 1 or y = - 26 x - 1

Solution steps

Slope of the tangent line at

(0, - 1) : m = \frac{3 x ^{2} + 2}{x}

Find the slope of

y = 3 x^{2} + 1 : m = 6 x

Solve

6 x = \frac{3 x ^{2} + 2}{x} : x = \frac{2}{3}, x = - \frac{2}{3}

Find the tangent line at

x = \frac{2}{3} : y = 26 x - 1

Find the tangent line at

x = - \frac{2}{3} : y = - 26 x - 1

y = 26 x - 1 or y = - 26 x - 1

y^{^{''}} - 7 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1}{8} x 64 - x^{2})

(5, - 4), (4, 7)

\int \frac{7}{x 36 - 49 ln ^{2} ( x )} d x

\frac{d}{d x} (x sin (x) + cos (x))

What is the tangent of y=3x^2+1,(0,-1) ?
The tangent of y=3x^2+1,(0,-1) is y=2sqrt(6)x-1 \lor y=-2sqrt(6)x-1