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

The tangent of f(x)=3x^2+4x,\at x=-2 is y=-8x-12

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tangent of f(x)=3x^2+4x,\at x=-2

tangent of

f (x) = 3 x^{2} + 4 x, at x = - 2

y = - 8 x - 12

Solution steps

Find the tangent point:

(- 2, 4)

Find the slope of

f (x) = 3 x^{2} + 4 x : \frac{df ( x )}{d x} = 6 x + 4

Compute the slope of

f (x) = 3 x^{2} + 4 x

x = - 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(- 2, 4) : y = - 8 x - 12

y = - 8 x - 12

x \to 0 lim (cos (\frac{7}{x}))

x \to 9 lim (81 - x^{2})

f (x) = - 8 x^{2}

y = 12 (ln (2 x) 3)

7 e^{2 t}

What is the tangent of f(x)=3x^2+4x,\at x=-2 ?
The tangent of f(x)=3x^2+4x,\at x=-2 is y=-8x-12