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

The tangent of f(x)=3x^2+4x,\at x=-3 is y=-14x-27

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

tangent of

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

y = - 14 x - 27

Solution steps

Find the tangent point:

(- 3, 15)

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 = - 3 : m = - 14

Find the line with slope m=

- 14

and passing through

(- 3, 15) : y = - 14 x - 27

y = - 14 x - 27

\int \frac{7}{2 y ^{4}} d y

y^{^{'}} = \frac{x ^{4} y ^{4} + 4 x ^{4}}{y ^{3}}

x^{3} + 2 x^{2} + 4

\int \frac{4 ^{x}}{x} d x

y^{^{''}} + 9 y = 4 sin (x)

What is the tangent of f(x)=3x^2+4x,\at x=-3 ?
The tangent of f(x)=3x^2+4x,\at x=-3 is y=-14x-27