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

The tangent of f(x)=x^4-13^2+36,\at x=-2 is y=-32x-181

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

tangent of

f (x) = x^{4} - 1 3^{2} + 36, at x = - 2

y = - 32 x - 181

Solution steps

Find the tangent point:

(- 2, - 117)

Find the slope of

f (x) = x^{4} - 1 3^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 32

Find the line with slope m=

- 32

and passing through

(- 2, - 117) : y = - 32 x - 181

y = - 32 x - 181

\frac{d}{d x} (arcsin (\frac{y}{x}))

\frac{\partial}{\partial x} (x y e^{x + 2 y})

\frac{d}{d x} (2 x - \frac{1}{x ^{2}})

\frac{d y}{d t}, y = e^{t}

\int sin^{5} (x) d x

What is the tangent of f(x)=x^4-13^2+36,\at x=-2 ?
The tangent of f(x)=x^4-13^2+36,\at x=-2 is y=-32x-181