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

The tangent of f(x)=3x^2-4x,\at x=-1 is y=-10x-3

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

tangent of

f (x) = 3 x^{2} - 4 x, at x = - 1

y = - 10 x - 3

Solution steps

Find the tangent point:

(- 1, 7)

Find the slope of

f (x) = 3 x^{2} - 4 x : \frac{df ( x )}{d x} = 6 x - 4

Compute the slope of

f (x) = 3 x^{2} - 4 x

x = - 1 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 1, 7) : y = - 10 x - 3

y = - 10 x - 3

\frac{d}{d y} (x - 2 y)

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

f (x) = 5 x^{9} x^{6}, at x = 2

y^{^{'}} = \frac{y ^{4} - 3}{4 x y ^{3}}

f (x) = x^{\frac{3}{5}}

What is the tangent of f(x)=3x^2-4x,\at x=-1 ?
The tangent of f(x)=3x^2-4x,\at x=-1 is y=-10x-3