What is the tangent of f(x)=2x^2+8x,\at x=2 ?

The tangent of f(x)=2x^2+8x,\at x=2 is y=16x-8

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tangent of f(x)=2x^2+8x,\at x=2

tangent of

f (x) = 2 x^{2} + 8 x, at x = 2

y = 16 x - 8

Solution steps

Find the tangent point:

(2, 24)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2} + 8 x

x = 2 : m = 16

Find the line with slope m=

16

and passing through

(2, 24) : y = 16 x - 8

y = 16 x - 8

\int \frac{2 X ^{2} - 7 X - 2}{X ^{3} - X} d X

y^{^{'}} = t^{0.5}, y (0) = 0

\frac{d}{d x} (5 x e^{3 x})

\int_{0}^{\frac{π}{3}} 8 sec^{3} (x) d x

f (x) = 7 x + 2, (2, 4)

What is the tangent of f(x)=2x^2+8x,\at x=2 ?
The tangent of f(x)=2x^2+8x,\at x=2 is y=16x-8