What is the tangent of y=-9x^2 ?

The tangent of y=-9x^2 is y=-18a_{0}x+9a_{0}^2

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tangent of y=-9x^2

tangent of

y = - 9 x^{2}

y = - 18 a_{0} x + 9 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 9 a_{0}^{2})

Find the slope of

y = - 9 x^{2} : \frac{d y}{d x} = - 18 x

Compute the slope of

y = - 9 x^{2}

x = a_{0} : m = - 18 a_{0}

Find the line with slope m=

- 18 a_{0}

and passing through

(a_{0}, - 9 a_{0}^{2}) : y = - 18 a_{0} x + 9 a_{0}^{2}

y = - 18 a_{0} x + 9 a_{0}^{2}

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 0 ^{n}}

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

\frac{d}{d x} (sin (3 x^{4}))

\frac{d}{d x} (6 sin (x) - 8 e^{x})

\frac{d ^{2}}{d x ^{2}} (\frac{1 - 5 x}{1 + 5 x})