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Mathematics 14 Online
OpenStudy (anonymous):

The net profit of a company for the years 2000 to 2011 is approximated by the function p(t) = t³ – 5tÆ + 6t, where p is the profit (in thousands of dollars) and t is the number of years that have passed since January 1, 2000. For which period will the company report a net loss?

OpenStudy (anonymous):

Can someone please help

jimthompson5910 (jim_thompson5910):

is that "Æ" symbol a typo?

jimthompson5910 (jim_thompson5910):

refresh if you get a bunch of weird symbols

OpenStudy (anonymous):

Its 5t^2 thats what its supposed to be

jimthompson5910 (jim_thompson5910):

oh ok, so \[\Large p(t) = t^3 - 5t^2 + 6t\] right?

OpenStudy (anonymous):

yes

jimthompson5910 (jim_thompson5910):

are you in calculus class?

OpenStudy (anonymous):

no. Im on a online math class (12th) grade

OpenStudy (anonymous):

Im taking math 3

jimthompson5910 (jim_thompson5910):

alright, then the best way to do this is to graph p(t) I recommend you use this graphing calculator https://www.desmos.com/calculator

jimthompson5910 (jim_thompson5910):

type in x^3 - 5x^2 + 6x into the calculator

OpenStudy (anonymous):

okay

jimthompson5910 (jim_thompson5910):

when is the curve below the x axis? for which x values? keep in mind that x >= 0

OpenStudy (anonymous):

its below the x-axis between the positive 2 and 3

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

x corresponds to the number of years after the year 2000

jimthompson5910 (jim_thompson5910):

since 2 < x < 3 is when p(t) is below the x axis, this means the company has a net loss between 2002 and 2003

OpenStudy (anonymous):

so the answer is ' January 1,2002-December 31,2002 ?

jimthompson5910 (jim_thompson5910):

well on Jan 1, 2002, the p(t) value is actually 0 this is because p(2) = 0. There is no net profit and no net loss the next day is when the losses start happening and they continue to happen until Jan 1, 2003. On this day, you break even again because p(3) = 0. Everything after t = 3 is a net profit

jimthompson5910 (jim_thompson5910):

So I'd say from Jan 2, 2002 to Dec 31, 2002 is when you'd have a net loss

OpenStudy (anonymous):

Thank you soooo much : ) I have one more question

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

Which statement about the graph of the rational expression x^2+6x+9 / x^2-4

jimthompson5910 (jim_thompson5910):

seems like an incomplete question

OpenStudy (anonymous):

function is above the x axis between -2 and 2. function is above the x axis and x is greater than 2. function is below the x axis for x is less than -2. function is below x axis between -3 and -2

OpenStudy (anonymous):

Do I use the graphing calculator?

jimthompson5910 (jim_thompson5910):

so graph (x^2+6x+9)/(x^2-4) using desmos and you should see the answer (or at least eliminate the nonanswers)

jimthompson5910 (jim_thompson5910):

yes you would

OpenStudy (anonymous):

How do I divide using the graphing calculator? I'm using the / symbol but it's making the 9 the numerator instead of dividing the whole equation

jimthompson5910 (jim_thompson5910):

you need to put parenthesis around the whole numerator

jimthompson5910 (jim_thompson5910):

same with the denominator

jimthompson5910 (jim_thompson5910):

(numerator)/(denominator)

jimthompson5910 (jim_thompson5910):

which is why you type in (x^2+6x+9)/(x^2-4)

OpenStudy (anonymous):

ok I used the ( )

jimthompson5910 (jim_thompson5910):

so what answers do you see? or what answers can you eliminate?

OpenStudy (anonymous):

We can eliminate anything below the x axis?

jimthompson5910 (jim_thompson5910):

lets go through the answer choices

jimthompson5910 (jim_thompson5910):

"function is above the x axis between -2 and 2" is that true or false?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

false

jimthompson5910 (jim_thompson5910):

why is it false

jimthompson5910 (jim_thompson5910):

which parts aren't above the x axis?

OpenStudy (anonymous):

I'm looking at it again. One moment

OpenStudy (anonymous):

Honestly, I'm not sure exactly what I'm looking at. Is it quadrant 3?

jimthompson5910 (jim_thompson5910):

well notice between x = -2 and x = 2 the entire portion of the graph is below the x axis not above so that's why A is false

OpenStudy (anonymous):

OK, that makes sense. Let me look at b.

jimthompson5910 (jim_thompson5910):

I'm assuming B is supposed to say "function is above the x axis when x is greater than 2." instead of "function is above the x axis and x is greater than 2."

OpenStudy (anonymous):

b would be true

jimthompson5910 (jim_thompson5910):

yes it is, when x > 2, the curve is above the x axis

jimthompson5910 (jim_thompson5910):

choice C "function is below the x axis for x is less than -2." true or false?

OpenStudy (anonymous):

true

jimthompson5910 (jim_thompson5910):

why true?

jimthompson5910 (jim_thompson5910):

is the curve below the x axis when you are to the left of x = -2 ?

OpenStudy (anonymous):

No it isn't. That's false.

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

finally choice D "function is below x axis between -3 and -2" true or false?

OpenStudy (anonymous):

false

OpenStudy (anonymous):

so the answer is b

jimthompson5910 (jim_thompson5910):

yes because the portion between x = -3 and x = -2 is above

jimthompson5910 (jim_thompson5910):

yes it is B

OpenStudy (anonymous):

THANKS FOR YOUR HELP!!!! : )

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

Can you help with another question?

jimthompson5910 (jim_thompson5910):

sure

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