What is the tangent of y=(x^2-2)(x^3-3x),(-2,-4) ?

The tangent of y=(x^2-2)(x^3-3x),(-2,-4) is y=26x+48

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tangent of y=(x^2-2)(x^3-3x),(-2,-4)

tangent of

y = (x^{2} - 2) (x^{3} - 3 x), (- 2, - 4)

y = 26 x + 48

Solution steps

Find the slope of

y = (x^{2} - 2) (x^{3} - 3 x) : \frac{d y}{d x} = 5 x^{4} - 15 x^{2} + 6

Compute the slope of

y = (x^{2} - 2) (x^{3} - 3 x)

(- 2, - 4) : m = 26

Find the line with slope m=

26

and passing through

(- 2, - 4) : y = 26 x + 48

y = 26 x + 48

\frac{d}{d x} (3 x^{2} - 4 x + 2)

y = x^{2} - 3 x + 3, (4, 7)

\frac{d ^{2}}{d x ^{2}} (4 x + 1)

\int 2 x + 5 e^{x} d x

\frac{d}{d x} (8 x - 1)

What is the tangent of y=(x^2-2)(x^3-3x),(-2,-4) ?
The tangent of y=(x^2-2)(x^3-3x),(-2,-4) is y=26x+48