The initial price of buzz.com stock is $20 per share. After 20 days the stock price is$30 per share and after 40 days the price is $35 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function.Find the multipart function s(t) giving the stock price after t days. {s(t) = (quadratic function) when 0≤t≤125 and t>125 --For this part I tried to form a quadratic equation out of the point (0,20) (20,30) (40,35). I know the stock price cannot equal zero, but it is unclear to me how/why it needs to be a parametric equation. the price must equal zero at 125 days? I think once I understand how to set up the equation I can do the other parts. but here they are: (a) If you buy 1000 shares after 30 days, what is the cost? (b) To maximize profit, when should you sell shares? How much will the profit be on your 1000 shares purchased in (a)? 2. ## Re: quadratic stock problem Originally Posted by UWstudent The initial price of buzz.com stock is$20 per share. After 20 days the stock price is $30 per share and after 40 days the price is$35 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function.Find the multipart function s(t) giving the stock price after t days.
{s(t) = (quadratic function) when 0≤t≤125 and t>125

--For this part I tried to form a quadratic equation out of the point (0,20) (20,30) (40,35). I know the stock price cannot equal zero, but it is unclear to me how/why it needs to be a parametric equation. the price must equal zero at 125 days?

I think once I understand how to set up the equation I can do the other parts. but here they are:

(a) If you buy 1000 shares after 30 days, what is the cost?
(b) To maximize profit, when should you sell shares?
How much will the profit be on your 1000 shares purchased in (a)?
a general quadratic is given by $a x^2 + b x + c$

we can plug your numbers in and get 3 equations in 3 unknowns, a, b, and c, which can be solved for.

if you do this you find that the stock price is modeled by an inverted parabola, i.e. a<0, that intersects the x axis at x=125.498.

I guess you would just say that the stock price is 0 after 125.5 days

Give a yell if you can't figure out the coefficients.

3. ## Re: quadratic stock problem

Originally Posted by UWstudent
The initial price of buzz.com stock is $20 per share. After 20 days the stock price is$30 per share and after 40 days the price is $35 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function.Find the multipart function s(t) giving the stock price after t days. {s(t) = (quadratic function) when 0≤t≤125 and t>125 --For this part I tried to form a quadratic equation out of the point (0,20) (20,30) (40,35). I know the stock price cannot equal zero, but it is unclear to me how/why it needs to be a parametric equation. the price must equal zero at 125 days? I think once I understand how to set up the equation I can do the other parts. but here they are: (a) If you buy 1000 shares after 30 days, what is the cost? (b) To maximize profit, when should you sell shares? How much will the profit be on your 1000 shares purchased in (a)? This question is worded VERY trickily. Actually, I hope it is worded better than you have said. Anyway, the first thing you need to understand is that the price of a stock cannot be less than zero. This is basically a fact of law. The second thing that you need to know is that any three distinct, non-co-linear points uniquely determine a quadratic equation. What was the equation that you computed from your three points? Does that equation apply on day 126 or thereafter? If so, why? If not, why not and what function does apply? 4. ## Re: quadratic stock problem I got a=-1/160 b=12.5/20 c=20 for the equation: -1/160t^2+0.625t+20=s(t) it is the same for both 0≤t≤125 and t>125 right? This answer was not accepted as correct, however. I've done it over and over. what is wrong? Thanks so much for your help!! 5. ## Re: quadratic stock problem The function would probably not apply after day 126, because the parabola is facing downward and after day 126 it is forever negative? I am only guessing because the equation I continue to get is not correct. 6. ## Re: quadratic stock problem Originally Posted by UWstudent I got a=-1/160 b=12.5/20 c=20 for the equation: -1/160t^2+0.625t+20=s(t) it is the same for both 0≤t≤125 and t>125 right? This answer was not accepted as correct, however. I've done it over and over. what is wrong? Thanks so much for your help!! Those coefficients are correct. The model parabola hits the x axis, i.e. the stock price becomes 0, at just below x=125.5. So at day 125, the model is valid. At day 126 it is not. If you are discretizing in days, I would write this as$sp(t)=\begin{cases} -\dfrac {x^2}{160}+\dfrac{25x}{40} + 20&t \leq 125 \\ \\ 0 &t>125 \end{cases}\$

so basically you are entirely correct.

I really can't give you advice from here on what format to put this in to make your software accept it.

7. ## Re: quadratic stock problem

Thank you so much!!!! Your formatting was correct. I spent sooooo long on this problem. I was beginning to think that I was incapable of doing precalculus.