Answer to Question #346428 in Calculus for bookaddict

Question

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.

Answer 1

a.

b.

$$y'=(-x^3+3x^2+6x-8)'=-3x^2+6x+6$$

c.

$$slope_1=m_1=y'(-2)$$

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

$$slope_2=m_2=y'(4)$$

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

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