What is the tangent of y=3x^3+x^2-8x+4,(-2,0) ?

The tangent of y=3x^3+x^2-8x+4,(-2,0) is y=24x+48

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tangent of y=3x^3+x^2-8x+4,(-2,0)

tangent of

y = 3 x^{3} + x^{2} - 8 x + 4, (- 2, 0)

y = 24 x + 48

Solution steps

Find the slope of

y = 3 x^{3} + x^{2} - 8 x + 4 : \frac{d y}{d x} = 9 x^{2} + 2 x - 8

Compute the slope of

y = 3 x^{3} + x^{2} - 8 x + 4

(- 2, 0) : m = 24

Find the line with slope m=

24

and passing through

(- 2, 0) : y = 24 x + 48

y = 24 x + 48

\int \frac{3 x}{7 - x ^{2}} d x

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

\int sin (θ) + cos (θ) d θ

x \to 5 lim (4)

x^{8 c o s (x)}

What is the tangent of y=3x^3+x^2-8x+4,(-2,0) ?
The tangent of y=3x^3+x^2-8x+4,(-2,0) is y=24x+48