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

The tangent of f(x)=4x^2-3x,\at x=-1 is y=-11x-4

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

tangent of

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

y = - 11 x - 4

Solution steps

Find the tangent point:

(- 1, 7)

Find the slope of

f (x) = 4 x^{2} - 3 x : \frac{df ( x )}{d x} = 8 x - 3

Compute the slope of

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

x = - 1 : m = - 11

Find the line with slope m=

- 11

and passing through

(- 1, 7) : y = - 11 x - 4

y = - 11 x - 4

x \to 2 lim (x^{2} + 2 x - 7)

\frac{\partial}{\partial y} (sin (2 y))

\int_{0}^{3} \frac{2 t}{( t - 4 ) ^{2}} d t

\frac{x ^{2} - 3 x + 2}{- x ^{2} - 4 x + 4}

\frac{\partial}{\partial x} (y^{4} sin (5 x))

What is the tangent of f(x)=4x^2-3x,\at x=-1 ?
The tangent of f(x)=4x^2-3x,\at x=-1 is y=-11x-4