What is the tangent of f(x)=9x^2,\at x=-2 ?

The tangent of f(x)=9x^2,\at x=-2 is y=-36x-36

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tangent of f(x)=9x^2,\at x=-2

tangent of

f (x) = 9 x^{2}, at x = - 2

y = - 36 x - 36

Solution steps

Find the tangent point:

(- 2, 36)

Find the slope of

f (x) = 9 x^{2} : \frac{df ( x )}{d x} = 18 x

Compute the slope of

f (x) = 9 x^{2}

x = - 2 : m = - 36

Find the line with slope m=

- 36

and passing through

(- 2, 36) : y = - 36 x - 36

y = - 36 x - 36

y = - x + 4, y = x^{2}, x = 0, x = 4

\int 3 x^{- 4} (2 + 4 x^{- 3})^{- 7} d x

f^{^{'}} (t) - f (t) = - 4 t

\int_{- 8}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

y^{^{'}} + 8 (tan (8 x)) y = - 2 cos (8 x)

What is the tangent of f(x)=9x^2,\at x=-2 ?
The tangent of f(x)=9x^2,\at x=-2 is y=-36x-36