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

The tangent of f(x)=x^2-5x,\at x=-2 is y=-9x-4

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

tangent of

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

y = - 9 x - 4

Solution steps

Find the tangent point:

(- 2, 14)

Find the slope of

f (x) = x^{2} - 5 x : \frac{df ( x )}{d x} = 2 x - 5

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 = - 9

Find the line with slope m=

- 9

and passing through

(- 2, 14) : y = - 9 x - 4

y = - 9 x - 4

\int 9 x sin (x) cos (x) d x

f (x) = ∣ x ∣ 5 - x^{2}, (2, 2)

x \to 2 lim (3 x + 4 x)

\int_{2}^{4} \frac{2}{x x ^{2} - 4} d x

x \to - 1 + lim (\frac{2}{x + 1})

What is the tangent of f(x)=x^2-5x,\at x=-2 ?
The tangent of f(x)=x^2-5x,\at x=-2 is y=-9x-4