Problem solving and reasoning

Question

The price of shares in a company named ‘Aspire’ rose and then fell during one day of trade. Trade began at 10 am and closer at 4pm. It was noted that the company’s share price followed the cure P= —0.1(t —5)² +3 where $P was the price of shares t hours after 10am.

(a) At what time was the price share at its highest ?

(b) What was the closing price for the shares at the end of trade?

Answer 1

P = -0.1(t-5)² + 3

 

The graph is a parabola opening downward with vertex (highest point) (5,3)

 

(a). Highest price was $3.  This price was reached when t = 5  (3 PM)

 

(b). closing price = P(6) = -0.1(6-5)² + 3 = $2.90