What is the tangent of f(x)=3x^2+5x,\at x=-3 ?

The tangent of f(x)=3x^2+5x,\at x=-3 is y=-13x-27

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tangent of f(x)=3x^2+5x,\at x=-3

tangent of

f (x) = 3 x^{2} + 5 x, at x = - 3

y = - 13 x - 27

Solution steps

Find the tangent point:

(- 3, 12)

Find the slope of

f (x) = 3 x^{2} + 5 x : \frac{df ( x )}{d x} = 6 x + 5

Compute the slope of

f (x) = 3 x^{2} + 5 x

x = - 3 : m = - 13

Find the line with slope m=

- 13

and passing through

(- 3, 12) : y = - 13 x - 27

y = - 13 x - 27

\frac{\partial}{\partial x} (e^{- 5 x})

f (x) = \frac{( 1 )}{1 + x ^{2}}, at x = 1

\frac{\partial}{\partial x} (\frac{1}{1 + 0 ^{2}})

2 ln (cos (x))

- \frac{y}{2}

What is the tangent of f(x)=3x^2+5x,\at x=-3 ?
The tangent of f(x)=3x^2+5x,\at x=-3 is y=-13x-27