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

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

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

tangent of

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

y = - 3 x - 4

Solution steps

Find the tangent point:

(- 2, 2)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 3

Find the line with slope m=

- 3

and passing through

(- 2, 2) : y = - 3 x - 4

y = - 3 x - 4

2 x \frac{d y}{d x} = y

x \to \frac{1}{3} lim (∣ 3 x - 1 ∣ - 5)

3 x^{2} y^{2} d x + (2 y^{3} x + 4 y^{3}) d y = 0

5 x^{^{''}} + 25 x^{^{'}} + 1700 x = 0

\frac{1}{y ^{\frac{1}{3}}} d y = 1 d x

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