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

The tangent of f(x)=2-3x^2,\at x=-2 is y=12x+14

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

tangent of

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

y = 12 x + 14

Solution steps

Find the tangent point:

(- 2, - 10)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = 12

Find the line with slope m=

12

and passing through

(- 2, - 10) : y = 12 x + 14

y = 12 x + 14

\frac{d y}{d x} = y^{2} - 2 y

\frac{d}{d x} (e^{2 x} \cdot sin (x^{2} + 5))

2 x^{3} - x^{2} + 3

\frac{d}{d y} (ln (x^{2} + y^{2}))

\frac{d y}{d x} = (x + y + 9)^{2}

What is the tangent of f(x)=2-3x^2,\at x=-2 ?
The tangent of f(x)=2-3x^2,\at x=-2 is y=12x+14