What is the tangent of f(x)=4x^2+2x,\at x=2 ?

The tangent of f(x)=4x^2+2x,\at x=2 is y=18x-16

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tangent of f(x)=4x^2+2x,\at x=2

tangent of

f (x) = 4 x^{2} + 2 x, at x = 2

y = 18 x - 16

Solution steps

Find the tangent point:

(2, 20)

Find the slope of

f (x) = 4 x^{2} + 2 x : \frac{df ( x )}{d x} = 8 x + 2

Compute the slope of

f (x) = 4 x^{2} + 2 x

x = 2 : m = 18

Find the line with slope m=

18

and passing through

(2, 20) : y = 18 x - 16

y = 18 x - 16

\frac{\partial}{\partial x} (sech^{2} (x - t))

\int x^{3} cos (x^{4} + 2) d x

\int_{0}^{3} \frac{3 t}{( t - 4 ) ^{2}} d t

\frac{d}{d x} (2 x + 7 + \frac{450}{x})

f (x) = \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}

What is the tangent of f(x)=4x^2+2x,\at x=2 ?
The tangent of f(x)=4x^2+2x,\at x=2 is y=18x-16