What is the tangent of-8x^2-2x ?

The tangent of-8x^2-2x is y=(-16a_{0}-2)x+8a_{0}^2

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tangent of-8x^2-2x

tangent of

- 8 x^{2} - 2 x

y = (- 16 a_{0} - 2) x + 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 8 a_{0}^{2} - 2 a_{0})

Find the slope of

y = - 8 x^{2} - 2 x : \frac{d y}{d x} = - 16 x - 2

Compute the slope of

y = - 8 x^{2} - 2 x

x = a_{0} : m = - 16 a_{0} - 2

Find the line with slope m=

- 16 a_{0} - 2

and passing through

(a_{0}, - 8 a_{0}^{2} - 2 a_{0}) : y = (- 16 a_{0} - 2) x + 8 a_{0}^{2}

y = (- 16 a_{0} - 2) x + 8 a_{0}^{2}

t \to 1 lim (\frac{e ^{t} - e}{t - 1})

\frac{d}{d x} (x^{2} - 16 I n (x^{2}))

n = 2 \sum \infty \frac{5}{6 ^{n}}

\int e^{x} - \frac{2}{x} d x

(x + y) y^{^{'}} = 1

What is the tangent of-8x^2-2x ?
The tangent of-8x^2-2x is y=(-16a_{0}-2)x+8a_{0}^2