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

The tangent of f(x)=x^2-3x,\at x=-2 is y=-7x-4

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

tangent of

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

y = - 7 x - 4

Solution steps

Find the tangent point:

(- 2, 10)

Find the slope of

f (x) = x^{2} - 3 x : \frac{df ( x )}{d x} = 2 x - 3

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 = - 7

Find the line with slope m=

- 7

and passing through

(- 2, 10) : y = - 7 x - 4

y = - 7 x - 4

1 - \frac{1}{x}

y^{'} = - \frac{2 y}{x} + 4 x

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

\frac{( x + 4 )}{x - 2}

x^{3} - x, 0

What is the tangent of f(x)=x^2-3x,\at x=-2 ?
The tangent of f(x)=x^2-3x,\at x=-2 is y=-7x-4