What is the tangent of f(x)=x^2+5x-2,\at x=-1 ?

The tangent of f(x)=x^2+5x-2,\at x=-1 is y=3x-3

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tangent of f(x)=x^2+5x-2,\at x=-1

tangent of

f (x) = x^{2} + 5 x - 2, at x = - 1

y = 3 x - 3

Solution steps

Find the tangent point:

(- 1, - 6)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 3

Find the line with slope m=

3

and passing through

(- 1, - 6) : y = 3 x - 3

y = 3 x - 3

x \to 0 + lim (\frac{6}{x ^{2} - x})

f (x) = x - 5

\frac{d}{d x} (3 x^{2} - x^{3 x})

\frac{\partial}{\partial x} ((2 x + 8 y)^{3})

\frac{d}{d x} (x^{- 1} ln (x))

What is the tangent of f(x)=x^2+5x-2,\at x=-1 ?
The tangent of f(x)=x^2+5x-2,\at x=-1 is y=3x-3