What is the tangent of f(x)=3x^2-4,\at x=2 ?

The tangent of f(x)=3x^2-4,\at x=2 is y=12x-16

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tangent of f(x)=3x^2-4,\at x=2

tangent of

f (x) = 3 x^{2} - 4, at x = 2

y = 12 x - 16

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

f (x) = 3 x^{2} - 4 : \frac{df ( x )}{d x} = 6 x

Compute the slope of

f (x) = 3 x^{2} - 4

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 8) : y = 12 x - 16

y = 12 x - 16

y^{^{''}} - y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

\int 8 + 7^{x} d x

f (x) = e^{5 x} cos (π x), at x = 1

\int - 2 e^{x} d x

\frac{d}{d x} (- 6 sin (2 x) + 3 cos (3 x))

What is the tangent of f(x)=3x^2-4,\at x=2 ?
The tangent of f(x)=3x^2-4,\at x=2 is y=12x-16