What is the tangent of-4x^2-3x+2,\at x=-1 ?

The tangent of-4x^2-3x+2,\at x=-1 is y=5x+6

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tangent of-4x^2-3x+2,\at x=-1

tangent of

- 4 x^{2} - 3 x + 2, at x = - 1

y = 5 x + 6

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

y = - 4 x^{2} - 3 x + 2 : \frac{d y}{d x} = - 8 x - 3

Compute the slope of

y = - 4 x^{2} - 3 x + 2

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

(- 1, 1) : y = 5 x + 6

y = 5 x + 6

\int \frac{1}{5 x ^{4}} d x

\frac{d}{d x} (\frac{sin ( x )}{1 + cos ( x )})

\int 22 x d x

\frac{d x}{d t} = \frac{e ^{t} - e ^{- t}}{3 + x}

y = \frac{1}{2} e^{2 x}

What is the tangent of-4x^2-3x+2,\at x=-1 ?
The tangent of-4x^2-3x+2,\at x=-1 is y=5x+6