What is the tangent of f(x)=x^3-2x+7,\at x=-2 ?

The tangent of f(x)=x^3-2x+7,\at x=-2 is y=10x+23

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tangent of f(x)=x^3-2x+7,\at x=-2

tangent of

f (x) = x^{3} - 2 x + 7, at x = - 2

y = 10 x + 23

Solution steps

Find the tangent point:

(- 2, 3)

Find the slope of

f (x) = x^{3} - 2 x + 7 : \frac{df ( x )}{d x} = 3 x^{2} - 2

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 = 10

Find the line with slope m=

10

and passing through

(- 2, 3) : y = 10 x + 23

y = 10 x + 23

x^{4} - 2 x^{2} + 5

n = 1 \sum \infty 8 e^{- 6 n}

\frac{y ^{^{'}} - e ^{- t} + 4}{y} = - 4

\int sec (x) tan (x) - 5 e^{x} d x

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

What is the tangent of f(x)=x^3-2x+7,\at x=-2 ?
The tangent of f(x)=x^3-2x+7,\at x=-2 is y=10x+23