What is the tangent of-2x^2,\at x=0 ?

The tangent of-2x^2,\at x=0 is y=0

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tangent of-2x^2,\at x=0

tangent of

- 2 x^{2}, at x = 0

y = 0

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = - 2 x^{2} : \frac{d y}{d x} = - 4 x

Compute the slope of

y = - 2 x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 0) : y = 0

y = 0

\frac{d}{d x} (\frac{2}{e ^{x} + \frac{1}{e ^{x}}})

\int \frac{1 - t ^{2}}{t ^{4}} d t

\int 10 ln (x^{2} - 1) d x

\frac{2 x - 6}{x + 1}, at x = 0

\frac{t}{t ^{2} + 36}