What is the tangent of f(x)=6x^2+6,\at x=-2 ?

The tangent of f(x)=6x^2+6,\at x=-2 is y=-24x-18

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tangent of f(x)=6x^2+6,\at x=-2

tangent of

f (x) = 6 x^{2} + 6, at x = - 2

y = - 24 x - 18

Solution steps

Find the tangent point:

(- 2, 30)

Find the slope of

f (x) = 6 x^{2} + 6 : \frac{df ( x )}{d x} = 12 x

Compute the slope of

f (x) = 6 x^{2} + 6

x = - 2 : m = - 24

Find the line with slope m=

- 24

and passing through

(- 2, 30) : y = - 24 x - 18

y = - 24 x - 18

(3 x - 1) (x - 2)^{4}

\int \frac{x ^{2}}{x ^{2} + 18} d x

x \to 3 lim (\frac{x - 4}{x ^{2} - 9})

\frac{d}{d x} ((2 x^{2} + 7)^{5})

\frac{\partial}{\partial x} (ln (e^{x} + x y^{3}))

What is the tangent of f(x)=6x^2+6,\at x=-2 ?
The tangent of f(x)=6x^2+6,\at x=-2 is y=-24x-18