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

The tangent of f(x)=4x-3x^2,\at x=2 is y=-8x+12

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

tangent of

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

y = - 8 x + 12

Solution steps

Find the tangent point:

(2, - 4)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 8

Find the line with slope m=

- 8

and passing through

(2, - 4) : y = - 8 x + 12

y = - 8 x + 12

d er i v a t i v e e^{, at} sin (b t)

\int csc (x) (csc (x) - cot (x)) d x

t an g e n t y = x^{3} - 2 x^{2} + 5, (2, 5)

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 8

\int \frac{x}{( x + 1 ) ^{2}} d x

What is the tangent of f(x)=4x-3x^2,\at x=2 ?
The tangent of f(x)=4x-3x^2,\at x=2 is y=-8x+12