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

The tangent of f(x)=x^2+x,\at x=-3 is y=-5x-9

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

tangent of

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

y = - 5 x - 9

Solution steps

Find the tangent point:

(- 3, 6)

Find the slope of

f (x) = x^{2} + x : \frac{df ( x )}{d x} = 2 x + 1

Compute the slope of

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

x = - 3 : m = - 5

Find the line with slope m=

- 5

and passing through

(- 3, 6) : y = - 5 x - 9

y = - 5 x - 9

y^{^{'}} + y = cos (t)

x \to - \infty lim (3)

\frac{d}{d x} (3^{s i n (3 x)})

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

e (t + 1)

What is the tangent of f(x)=x^2+x,\at x=-3 ?
The tangent of f(x)=x^2+x,\at x=-3 is y=-5x-9