What is the tangent of f(x)=x^3-2x^2-3x,(1,-4) ?

The tangent of f(x)=x^3-2x^2-3x,(1,-4) is y=-4x

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tangent of f(x)=x^3-2x^2-3x,(1,-4)

tangent of

f (x) = x^{3} - 2 x^{2} - 3 x, (1, - 4)

y = - 4 x

Solution steps

Find the slope of

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

Compute the slope of

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

(1, - 4) : m = - 4

Find the line with slope m=

- 4

and passing through

(1, - 4) : y = - 4 x

y = - 4 x

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

f (x) = 7 x^{2}

(- 5, 2), (5, 6)

t^{\frac{1}{2}}

\frac{d}{d x} (tan (\frac{y}{x}))

What is the tangent of f(x)=x^3-2x^2-3x,(1,-4) ?
The tangent of f(x)=x^3-2x^2-3x,(1,-4) is y=-4x