Question #346428

The curve 𝑦=−𝑥3+3𝑥2+6𝑥−8 cuts the 𝑥-axis at 𝑥=−2,𝑥=1 and 𝑥=4.

a. Sketch the curve, showing clearly the intersection with the coordinate axes.

b. Differentiate 𝑦=−𝑥3+3𝑥2+6𝑥−8

c. Show that the tangents to the curve at 𝑥=−2 and 𝑥=4 are parallel.


1
Expert's answer
2022-05-31T12:50:34-0400

a.



b.


y=(x3+3x2+6x8)=3x2+6x+6y'=(-x^3+3x^2+6x-8)'=-3x^2+6x+6

c.


slope1=m1=y(2)slope_1=m_1=y'(-2)

=3(2)2+6(2)+6=18=-3(-2)^2+6(-2)+6=-18


slope2=m2=y(4)slope_2=m_2=y'(4)

=3(4)2+6(4)+6=18=-3(4)^2+6(4)+6=-18

Since m1=18=m2,m_1=-18=m_2, then the tangents to the curve at 𝑥=2𝑥=−2 and 𝑥=4𝑥=4 are parallel.


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