Question

PLEASE HELP WILL GIVE MEDAL AND WILL FAN

The following function represents the profit P(n), in dollars, that a concert promoter makes by selling tickets for n dollars each:

P(n) = -250n^2 + 3,250n - 9,000

Part A: What are the zeroes of the above function, and what do they represent? Show your work. (4 points)

Part B: Find the maximum profit by completing the square of the function P(n). Show the steps of your work. (4 points)

Part C: What is the axis of symmetry of the function P(n)? (2 points)

Answer 1

@jdoe0001 @pgpilot326

Answer 2

hmm have you done quadratic factoring I assume?

like say factoring \(\bf x^2+4x+4\to (\qquad ?)(\qquad? )\)

Answer 3

yes

Answer 4

(x+2)(x+2)

Answer 5

right

Answer 6

so part A is just asking for the zeros of the function
or "solution" or "roots"

which you can get by just setting P(n) to 0, that is \(\bf P(n) = -250n^2 + 3,250n - 9,000\implies 0= -250n^2 + 3,250n - 9,000
\\ \quad \\
simplified\implies 0=-250(n^2+13n-36)\implies 0=n^2+13n-36\)

Answer 7

and you can get those by just factoring the quadratic

Answer 8

hmm actually wait a sec

Answer 9

I took out the -250 as common factor... need to change the signs

Answer 10

\(\bf P(n) = -250n^2 + 3,250n - 9,000\to 0= -250n^2 + 3,250n - 9,000
\\ \quad \\
simplified\implies 0=-250(n^2-13n+36)\implies 0=n^2-13n+36\)

anyhow, you can also factor that

Answer 11

when i factored it i got (n−9)(n−4) so would the 0s be -9 and -4?

Answer 12

yeap

Answer 13

welll hold .. not quite... the roots are n-9 and n-4 yes
the value for "n" would be

\(\large 0=(n-9)(n-4)\to
\begin{cases}
0=n-9\to &9=n
\\ \quad \\
0=n-4\to &4=n
\end{cases}\)

Answer 14

you know what a "perfect square trinomial" is, right?

Answer 15

yes

Answer 16

for part B

\(\bf 0=n^2-13n+36\implies 0=(n^2-13n)+36
\\ \quad \\
0=(n^2-13n+{\color{red}{ \square }}^2)+36\)

any ideas what our missing number there is, to get a "perfect square trinomial" ?

Answer 17

3?

Answer 18

i tried finding the maximum and i got 6.5 is that correct?

Answer 19

well... let us notice that the middle term of a perfect square trinomial is
2 * left term * right term

so one could say that

yeap... 6.5 is correct

\(\bf 13x=2\cdot x\cdot \square \implies \cfrac{13\cancel{ x }}{2\cancel{ x }}=\square \)

Answer 20

Would the axis of symmetry be 250?

Answer 21

well dunno yet on the axis, that'd depend on part B though

Answer 22

wait isn't part b done already because we determined that the maximum is 6.5 and thats what part b is asking for?

Answer 23

well.. not quite

Answer 24

so, keep in mind that, all we're doing is borrowing from our good friend Mr Zero, 0

so if we ADD 13/2, we have to also SUBTRACT 13/2

thus \(\bf 0=n^2-13n+36\implies 0=(n^2-13n)+36
\\ \quad \\
0=\left[n^2-13n+\left({\color{red}{ \frac{13}{2} }}\right)^2\right]+36-\left({\color{red}{ \frac{13}{2} }}\right)^2
\\ \quad \\
0=\left[n-\left(\frac{13}{2}\right)\right]^2+36-\frac{13}{2}\implies 0=\left[n-\left(\frac{13}{2}\right)\right]^2+\frac{59}{2}\)

Answer 25

?

Answer 26

heheh

Answer 27

that's what part B was asking, to "complete the square"

so you can find the maximum profit

Answer 28

so whats the answer for part b? 0=[n−(13/2)]^2+59/2

Answer 29

so the vertex of that parabola is at \(\bf 0=\left[n-{\color{blue}{ \left(\frac{13}{2}\right)}}\right]^2{\color{blue}{ +\frac{59}{2}}}\qquad vertex \ \left({\color{blue}{ \frac{13}{2}, \frac{59}{2}}}\right)\)

Answer 30

so the vertex is (6.5,29.5)

Answer 31

recall that before simplifying, it had a negative leading term coefficient
that means

the parabola is going downwards so yes, let us use (6.5, 29.5)

Answer 32

which means that the axis of symmetry is 6.5

Answer 33

yeap, the axis of symmetry is \(\bf x = 6.5\)