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

The tangent of f(x)=2x^2,\at x=-3 is y=-12x-18

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

tangent of

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

y = - 12 x - 18

Solution steps

Find the tangent point:

(- 3, 18)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2}

x = - 3 : m = - 12

Find the line with slope m=

- 12

and passing through

(- 3, 18) : y = - 12 x - 18

y = - 12 x - 18

\int csc^{2} (6 x + 3) d x

\frac{d}{d x} ((\frac{x}{2})^{4})

\frac{d}{d x} (a ln (2 x + b))

f (x) = (ln (x))^{3}

What is the tangent of f(x)=2x^2,\at x=-3 ?
The tangent of f(x)=2x^2,\at x=-3 is y=-12x-18