What is the tangent of 10x^2+2x ?

The tangent of 10x^2+2x is y=(20a_{0}+2)x-10a_{0}^2

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tangent of 10x^2+2x

tangent of

10 x^{2} + 2 x

y = (20 a_{0} + 2) x - 10 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 10 a_{0}^{2} + 2 a_{0})

Find the slope of

y = 10 x^{2} + 2 x : \frac{d y}{d x} = 20 x + 2

Compute the slope of

y = 10 x^{2} + 2 x

x = a_{0} : m = 20 a_{0} + 2

Find the line with slope m=

20 a_{0} + 2

and passing through

(a_{0}, 10 a_{0}^{2} + 2 a_{0}) : y = (20 a_{0} + 2) x - 10 a_{0}^{2}

y = (20 a_{0} + 2) x - 10 a_{0}^{2}

\int 8 t^{2} d t

(3 x - 4)^{3}

f (x) = (\frac{3}{4}) x^{\frac{4}{3}}

x \to - 8 lim (e (\frac{3}{( x - 8 )}))

\int \frac{x}{3 x ^{2} - 16} d x

What is the tangent of 10x^2+2x ?
The tangent of 10x^2+2x is y=(20a_{0}+2)x-10a_{0}^2