Completing the Square

We can identify \(d = 3\) and \(e = 4\). As a result, we can identify the vertex as \((3, 4)\)

Before graphing, we can create a table of values to identify the main points:

x Values	0	1	2	3	4	5	6
y Values	22	12	6	4	6	12	22

We can then sketch the corresponding graph as such:

As \(a = 1.5\), we can factor this value out of the first 2 terms:

\(y = 1.5(x^2 + 4x) - 5 \)

Next, as \(b = 4\), we can use this value to find the special number:

\(\left(\cfrac{b}{2}\right)^2 = \left(\cfrac{4}{2}\right)^2 = \cfrac{16}{4} = 4\)

We can add the special number to the brackets and subtract the expression by the special number multiplied by factor \(a\). We can then simplify the constants:

\(y = 1.5(x^2 + 4x + 4) - 5 - (1.5)(4)\)

\(y = 2(x^2 + 4x + 4) - 5 - 6\)

\(y = 2(x^2 + 4x + 4) - 11\)

We can factor the trinomial inside the brackets by finding \(2\) numbers that result in the sum of the second term and the product of the third term:

\(y = 1.5(x^2 + 2x + 2x + 4) - 11\)

\(y = 1.5(x(x + 2) + 2(x + 2)) - 11\)

\(y = 1.5(x + 2)^2 - 11\)

Therefore, we can determine that completing the square of function \(y = 1.5x^2 + 6x - 5\) results in \(1.5(x + 2)^2 - 11 = 0\).

We can also determine that the function has a minimum at \((-2, -11)\) since \(a\) is positive.

Before graphing, we can create a table of values to identify the main points:

x Values	-5	-4	-3	-2	-1	0	1
y Values	2.5	-5	-9.5	-11	-9.5	-5	2.5

We can now sketch our graph:

i. First, we can use the Revenue formula to describe the ticket price:

\(\text{Revenue} = \text{Price} \times \text{Quantity}\)

OR

\(R = P \times Q\)

Next, we can determine the respective expressions for Price and Quantitiy and combine them to determine the Revenue formula:

\(P = 20 + 2x\)

\(Q = 1200 - 60x\)

\(R = (20 + 2x)(1200 - 60x)\)

Therefore, we can determine that the revenue revelation describing the ticket sales is \(R = (20 + 2x)(1200 - 60x)\).

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ii. In order to find the maximum revenue (or max point) for the ticket sales, we must complete the square.
We can expand the expression using distributive property, collect like terms and rearrange in descending order:

\(R = (20)(1200) + (20)(-60x) + (2x)(1200) + (2x)(-60x)\)

\(R = 24000 - 1200x + 2400x - 120x^2\)

\(R = -120x^2 + 1200x + 24000\)

Next, as \(a = -120\), we can factor that value out of the first \(2\) terms:

\(R = -120(x^2 - 10x) + 24000\)

Then, as \(b = -10\), we can use this value to determine the special number:

\(\left(\cfrac{b}{2}\right)^2 = \left(\cfrac{-10}{2}\right)^2 = \cfrac{100}{4} = 25\)

We can add the special number to the brackets and subtract the expression by the special number multiplied by factor \(a\). We can then simplify the constants:

\(R = -120(x^2 - 10x + 25) + 24000 - (-120)(25)\)

\(R = -120(x^2 - 10x + 25) + 24000 + 3000\)

\(R = -120(x^2 - 10x + 25) + 27000\)

We can factor the trinomial inside the brackets by finding \(2\) numbers that result in the sum of the second term and the product of the third term:

\(R = -120(x^2 - 5x - 5x + 25) + 27000\)

\(R = -120(x(x - 5) -5(x - 5)) + 27000\)

\(R = -120(x - 5)^2 + 27000\)

As \(x = 5\), we can plug this value into the Price formula to determine what the new ticket prices are:

\(P = 20 + 2x\)

\(P = 20 + (2)(5)\)

\(P = $30\)

Therefore, we can determine that the selling price to maximize revenue is \($30/[\text{ticket}]\).

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iii. We can use \(x = 5\) to determine the maximum amount of tickets that will be sold using the Quantity formula:

\(Q = 1200 - 60x\)

\(Q = 1200 - (60)(5)\)

\(Q = 900\;[\text{tickets}]\)

Therefore, we can determine that we can sell \(900\; [\text{tickets}]\) at an adjusted price of \($30\).

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iv. As \(x = 5\), we can plug this value into the Revenue formula to determine what the new ticket prices are:

\(R = (20 + 2x)(1200 - 60x)\)

\(R = (20 + (2)(5))(1200 - (60)(5))\)

\(R = (20 + 10)(1200 - 300)\)

\(R = (30)(900)\)

\(R = $27000\)

Therefore, we can determine that the max revenue will be \($27000\).