What is the tangent of f(x)=2x^3+5x^2-2,\at x=-2 ?

The tangent of f(x)=2x^3+5x^2-2,\at x=-2 is y=4x+10

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tangent of f(x)=2x^3+5x^2-2,\at x=-2

tangent of

f (x) = 2 x^{3} + 5 x^{2} - 2, at x = - 2

y = 4 x + 10

Solution steps

Find the tangent point:

(- 2, 2)

Find the slope of

f (x) = 2 x^{3} + 5 x^{2} - 2 : \frac{df ( x )}{d x} = 6 x^{2} + 10 x

Compute the slope of

f (x) = 2 x^{3} + 5 x^{2} - 2

x = - 2 : m = 4

Find the line with slope m=

4

and passing through

(- 2, 2) : y = 4 x + 10

y = 4 x + 10

\frac{( 3 t - 2 ) ^{3}}{t + 5}

\frac{( e ^{- x} )}{x}

\frac{d}{d x} (cos (arccos (a x^{2} + b x + c)))

\frac{\partial}{\partial x} ((e^{x} - 4 x) sin (y))

y = (x - 2) (x^{2} + 4), (1, - 5)

What is the tangent of f(x)=2x^3+5x^2-2,\at x=-2 ?
The tangent of f(x)=2x^3+5x^2-2,\at x=-2 is y=4x+10