What is the tangent of 4x^2+2x,\at-3 ?

The tangent of 4x^2+2x,\at-3 is y=(8a_{0}+2)x-4a_{0}^2

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tangent of 4x^2+2x,\at-3

tangent of

4 x^{2} + 2 x, at - 3

y = (8 a_{0} + 2) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 2 a_{0})

Find the slope of

y = 4 x^{2} + 2 x : \frac{d y}{d x} = 8 x + 2

Compute the slope of

y = 4 x^{2} + 2 x

x = a_{0} : m = 8 a_{0} + 2

Find the line with slope m=

8 a_{0} + 2

and passing through

(a_{0}, 4 a_{0}^{2} + 2 a_{0}) : y = (8 a_{0} + 2) x - 4 a_{0}^{2}

y = (8 a_{0} + 2) x - 4 a_{0}^{2}

\int \frac{r}{r ^{2} + z ^{2}} d r

\frac{d}{d x} (\frac{1}{1 + \frac{1}{x}})

x \to 3 lim (x^{2} + 5 x)

\frac{d y}{d x} = \frac{1}{11} y cos^{2} (y)

\frac{d y}{d x} = - 9 x - 9 y

What is the tangent of 4x^2+2x,\at-3 ?
The tangent of 4x^2+2x,\at-3 is y=(8a_{0}+2)x-4a_{0}^2