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

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

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

tangent of

y = 3 x^{2} - 2 x, at x = - 1

y = - 8 x - 3

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

y = 3 x^{2} - 2 x : \frac{d y}{d x} = 6 x - 2

Compute the slope of

y = 3 x^{2} - 2 x

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}{d x} (x^{2} (x - 2)^{4})

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

x \to \infty lim (\frac{3}{3})

\frac{d}{d x} (x^{l n (5)})

n = 0 \sum \infty \frac{1}{n ^{2} + 5 n + 6}

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