What is the tangent of 5x^2-2x ?

The tangent of 5x^2-2x is y=(10a_{0}-2)x-5a_{0}^2

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tangent of 5x^2-2x

tangent of

5 x^{2} - 2 x

y = (10 a_{0} - 2) x - 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - 2 a_{0})

Find the slope of

y = 5 x^{2} - 2 x : \frac{d y}{d x} = 10 x - 2

Compute the slope of

y = 5 x^{2} - 2 x

x = a_{0} : m = 10 a_{0} - 2

Find the line with slope m=

10 a_{0} - 2

and passing through

(a_{0}, 5 a_{0}^{2} - 2 a_{0}) : y = (10 a_{0} - 2) x - 5 a_{0}^{2}

y = (10 a_{0} - 2) x - 5 a_{0}^{2}

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{8} e^{- x} d y = 0

\frac{d}{d x} (- a^{2} sin (a x))

\frac{d}{d t} (e^{6 t})

\int x (5 x^{2} + 9)^{8} d x

y = e^{x^{5}}

What is the tangent of 5x^2-2x ?
The tangent of 5x^2-2x is y=(10a_{0}-2)x-5a_{0}^2