Answer to Question #346428 in Calculus for bookaddict

Question #346428

The curve 𝑦=βˆ’π‘₯3+3π‘₯2+6π‘₯βˆ’8 cuts the π‘₯-axis at π‘₯=βˆ’2,π‘₯=1 and π‘₯=4.

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

b. Differentiate 𝑦=βˆ’π‘₯3+3π‘₯2+6π‘₯βˆ’8

c. Show that the tangents to the curve at π‘₯=βˆ’2 and π‘₯=4 are parallel.


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

a.



b.


yβ€²=(βˆ’x3+3x2+6xβˆ’8)β€²=βˆ’3x2+6x+6y'=(-x^3+3x^2+6x-8)'=-3x^2+6x+6

c.


slope1=m1=yβ€²(βˆ’2)slope_1=m_1=y'(-2)

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


slope2=m2=yβ€²(4)slope_2=m_2=y'(4)

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

Since m1=βˆ’18=m2,m_1=-18=m_2, then the tangents to the curve at π‘₯=βˆ’2π‘₯=βˆ’2 and π‘₯=4π‘₯=4 are parallel.


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