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

The tangent of f(x)=2x-3x^2,\at x=-1 is y=8x+3

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

tangent of

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

y = 8 x + 3

Solution steps

Find the tangent point:

(- 1, - 5)

Find the slope of

f (x) = 2 x - 3 x^{2} : \frac{df ( x )}{d x} = 2 - 6 x

Compute the slope of

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

x = - 1 : m = 8

Find the line with slope m=

8

and passing through

(- 1, - 5) : y = 8 x + 3

y = 8 x + 3

\frac{d y}{d t} = y + 2 t

- 19.6 t + 2

f (x) = sin (x^{4})

\int (9 x^{2} - 4 x + 2) d x

\frac{4}{( s + 1 ) ( s ^{2} + 4 s + 8 )}

What is the tangent of f(x)=2x-3x^2,\at x=-1 ?
The tangent of f(x)=2x-3x^2,\at x=-1 is y=8x+3