Question

We are given a model of passenger movements through an airport as: `y=592x^2-1002x+89401` where; `y` is passenger movement in thousands and `x` is years after 2008. Find (a) the passenger movement in 2015 and (b) the year passenger movements exceed 150m.?

Answer 1

a. 111,395 passengers
b. 2019

Explanation:

Okay, let's start off with part a.

`x` is the number of years after 2008

`y` is the number of passengers, in thousands

If we want the number of passengers in 2015, we first need to find `x`, the number of years after 2008.

`2015-2008=7` years

Now, we can plug it into the equation:

`y=592x^2−1002x+89,401`

`y=592*7^2-1002*7+89,401`

(Obviously, you will need a calculator for this part!)

`y=111,395` thousand passengers.

Let's move on to part b. Now, the problem gives us the value of `y` (150,000 passengers), and we must figure out `x` (the number of years after 2008):

`y=592x^2−1002x+89,401`

`150,000=592x^2-1002x+89,401`

`0=592x^2-1002x-60,599`

From here, there are three ways to solve this quadratic equation. You can factor it (too hard), use the quadratic equation (too hard), or graph it and find the zeros (complicated, but the best choice).

Here is our graph of `y=592x^2-1002x-60,599`

graph{592x^2-1002x-60599 [-114.3, 120, -52.8, 64.3]}

Obviously, there is more to this graph, but we only care about when the function is equal to 0 (when it crosses the x-axis). We can see there are two intersections, `(-9.3,0)` and `(11,0)`. Since the answer asks for a year after 2008, we can get rid of the negative value. Therefore, the number of passengers will reach 150,000 after 11 years. That year will be:

`2008+11=2019`

Hope this helps!

Answer 2

See solutions below.

Explanation:

We are given the model: `y=592x^2-1002x+89401`
Where; `y` is passenger movement in thousands
and `x` is years after 2008.

(a) To find the passenger movement in 2015;

`x= 2015 - 2008 =7`

Hence, `y_2015 = 592xx 7^2-1002xx 7+89401`

`= 29,008-7,014+89,401 = 111,395` thousand [Answer as provided]

(b) To find the year in which passenger movements exceed 150,000 thousand.

To find our year from 2008 when this occurs we could simply set `y` to 150,000 to find the offset. However, this would leave us with some rather nasty arithmetic. Since we are looking to an integer offset, we do not need the model to be as accurate as it is.

Let's round the model to hundreds of thousands by dividing through by 100 and rounding the coefficients.

`y/100 approx 6x^2 = 10x + 894`

To find the offset when passenger movement reaches 150,000k

`1500 approx 6x^2-10x+894`

For the pursose of our model we can assume this to be an equality.

Hence: `6x^2-10x+894 - 1500 =0`

Applying the quadratic formula

`x = (10+-sqrt(10^2+4xx6xx606))/12`

`= (10+-sqrt(100+14544))/12`

`approx(10+-121.01)/12`

We are only interested in the positive result.

`x approx 11` [To nearest integer]

Hence, our offset from 2008 is 11 years.

The passenger movements will exceed 150,000k in year
`2008+11= 2019` [Answer as provided]