Parabolas Application

Question

The price of shares in fledging company 'Lollies 'r us' plunged dramitically one afternoon, the following the breakout of a small fire on the premises. However, Ms Sarah Sayva of Lollies Anonymous agreed to back the company, and share prices began to rise. Sarah noted at the close of trade that afternoon that the company's share price followed the curve: P=0.1 (t-3)2 +1 where the $P is the price of the shares t hours afternoon.

 

a)What was the initial share price?

b)What was the lowest price of shares that afternoon? 

c) At what time was the price at its lowest?

d) What was the final price of 'Lollies 'r us' shares as trade closed at 5 pm?

e) Sketch a graph of the relationship between time and share price to represent the situation

 

I really need helps. Its parabola application and needs working out.

Answer 1

Hi! What Sarah is seeing is called the "vertex form" of a parabola. The vertex, or central point of the parabola, is located at (h,k), where the vertex form gives h and k in the form: P(t) = a(x - h)^2 + k where a is a scale factor, and (h,k) is the central point, or vertex, of the parabola. The particular parabola that Sarah sees follows this form. The constants that are used to make it are: a = 0.1 h = 3 k = 1 We use these to solve for the things the question asks us to calculate. Assume the fire occurred at noon, and set t=0 at that time. From the coefficients we've written above, we can note: P(t = 0) = 0.1(0 - 3)^2 + 1 = 0.1(3)^2 + 1 = 0.9 + 1 = $1.90 = initial price, P(t = 3) = 0.1 (3 - 3)^2 + 1 = 0.1 (0)^2 + 1 = $1.00 = lowest price, Lowest price was at t = 3 hours after the fire, P(t = 5) = 0.1(5 - 3)^2 + 1 = 0.1(2)^2 + 1 = $1.40 = closing price at 5 pm The graph would look like a parabola with its lowest point (its vertex) at (3,1) on a graph. The original share price of $1.90 would be the value of the stock at t=0, and would be the y-intercept of the graph. Best, Joe