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

The tangent of f(x)=2x^2+3x-2,\at x=-1 is y=-x-4

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

tangent of

f (x) = 2 x^{2} + 3 x - 2, at x = - 1

y = - x - 4

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

f (x) = 2 x^{2} + 3 x - 2 : \frac{df ( x )}{d x} = 4 x + 3

Compute the slope of

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

x = - 1 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 4

\frac{d}{d x} (k (1 - \frac{w}{d} e^{- s x} sin (d x + t)))

- x

\frac{9 - x ^{2}}{16 x}

\int π x d x

\frac{d}{d x} (i n^{2} (x + 3))

What is the tangent of f(x)=2x^2+3x-2,\at x=-1 ?
The tangent of f(x)=2x^2+3x-2,\at x=-1 is y=-x-4