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

The tangent of f(x)=3x^2+2x,\at x=-2 is y=-10x-12

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

tangent of

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

y = - 10 x - 12

Solution steps

Find the tangent point:

(- 2, 8)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 2, 8) : y = - 10 x - 12

y = - 10 x - 12

\frac{d}{d t} (sin (6 t))

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

y = 3 x^{2}, y = 6 - 3 x

y = x, y = 5 x, y = - x + 4

\int \frac{2 x + 1}{x ^{2} + 2 x + 17} d x

What is the tangent of f(x)=3x^2+2x,\at x=-2 ?
The tangent of f(x)=3x^2+2x,\at x=-2 is y=-10x-12