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

The tangent of f(x)=-4x^2-3+2,\at x=-1 is y=8x+3

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

tangent of

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

y = 8 x + 3

Solution steps

Find the tangent point:

(- 1, - 5)

Find the slope of

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

Compute the slope of

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

\int_{0}^{1} \frac{x + 1}{x ^{2} + 1} d x

\int \frac{e ^{\frac{1}{x ^{8}}}}{x ^{9}} d x

x \to - 3 lim (\frac{x - 3}{9 x})

\frac{\partial}{\partial x} (x^{2} + 5 x z + 3 y z^{2})

f (x) = 8 ln (t^{8})

What is the tangent of f(x)=-4x^2-3+2,\at x=-1 ?
The tangent of f(x)=-4x^2-3+2,\at x=-1 is y=8x+3