What is the tangent of f(x)=2x^3+3x^2-12x+4,\at x=0 ?

The tangent of f(x)=2x^3+3x^2-12x+4,\at x=0 is y=-12x+4

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tangent of f(x)=2x^3+3x^2-12x+4,\at x=0

tangent of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4, at x = 0

y = - 12 x + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4 : \frac{df ( x )}{d x} = 6 x^{2} + 6 x - 12

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 = - 12

Find the line with slope m=

- 12

and passing through

(0, 4) : y = - 12 x + 4

y = - 12 x + 4

\frac{\partial}{\partial x} (\frac{1}{( x - y )})

\int e^{x + 9 e^{x}} d x

y = 2 x^{\frac{1}{3}}, 0 \leq x \leq 32

4 (3 x + y - 2) d x - (3 x + y) d y = 0

\frac{d}{d x} (arcsin (x) - 1 - x^{2})

What is the tangent of f(x)=2x^3+3x^2-12x+4,\at x=0 ?
The tangent of f(x)=2x^3+3x^2-12x+4,\at x=0 is y=-12x+4