What is the tangent of f(x)=x^4-5x^2+4,\at x=-2 ?

The tangent of f(x)=x^4-5x^2+4,\at x=-2 is y=-12x-24

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tangent of f(x)=x^4-5x^2+4,\at x=-2

tangent of

f (x) = x^{4} - 5 x^{2} + 4, at x = - 2

y = - 12 x - 24

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

f (x) = x^{4} - 5 x^{2} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 10 x

Compute the slope of

f (x) = x^{4} - 5 x^{2} + 4

x = - 2 : m = - 12

Find the line with slope m=

- 12

and passing through

(- 2, 0) : y = - 12 x - 24

y = - 12 x - 24

y = sin (3) x^{2}

\int (ln (x) + x) d x

\frac{1}{( ( 25 s ^{2} + 19980 ) )}

\frac{\partial}{\partial x} (\frac{1}{x} sin (\frac{y}{x}))

\int (x - 1) sin (π) x d x

What is the tangent of f(x)=x^4-5x^2+4,\at x=-2 ?
The tangent of f(x)=x^4-5x^2+4,\at x=-2 is y=-12x-24