What is the tangent of f(x)=x^4-13x^2+36,\at x=-2 ?

The tangent of f(x)=x^4-13x^2+36,\at x=-2 is y=20x+40

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tangent of f(x)=x^4-13x^2+36,\at x=-2

tangent of

f (x) = x^{4} - 13 x^{2} + 36, at x = - 2

y = 20 x + 40

Solution steps

Find the tangent point:

(- 2, 0)

Find the slope of

f (x) = x^{4} - 13 x^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3} - 26 x

Compute the slope of

f (x) = x^{4} - 13 x^{2} + 36

x = - 2 : m = 20

Find the line with slope m=

20

and passing through

(- 2, 0) : y = 20 x + 40

y = 20 x + 40

\frac{d}{d x} (5 x^{3} - 5 x)

\frac{\partial}{\partial x} (cos (0.6 x - t))

\int \frac{300}{2 x + 25} d x

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

x \to - 2 lim (2 x^{3} - x^{2} + 7 x - 3)

What is the tangent of f(x)=x^4-13x^2+36,\at x=-2 ?
The tangent of f(x)=x^4-13x^2+36,\at x=-2 is y=20x+40