IMPORTANCE
OF
2017/12/11
Leo Yu
QUADRATIC
MODELS
SITUATION 1
CREATE A QUESTION THAT IS RELEVANT TO THE
SITUATION
THE PROBLEM
Your home need a frame.
The width of the frame is same.
The area should be 28cm2.
The inside of the frame must be 11 cm by
6 cm.
Determine the width of the frame.
PROVIDE A SOLUTION
TO YOUR QUESTION
THE QUADRATIC RELATION
BEFORE
y = (11 + 2x) × (6 + 2x)
y = 66 + 22x + 12x + 4x 2
y = 4x 2 + 34x + 66
AFTER
y = 4x 2 + 34x + 66 − 66
y = 4x 2 + 34x
x
DESCRIBE HOW THE SITUATION REPRESENTS A
QUADRATIC RELATION
Before the description:
This is rectangle 1 and 2:
Y is the area before and after. The width of the rectangle 1 is (6 + 2x) and the length
of the rectangle is (11 + 2x). The area can be a
Click the play button to see
quadratic relation.
the graph
So, the area of rectangle 1 is (6 + 2x) × (11 + 2x). the
standard form is 4x 2 + 34x + 66.
The area of the frame is (the area of rectangle 1) – (the
area of rectangle 2).
So, it is 4x 2 + 34x + 66– 11 × 6 = 4x 2 + 34x.
The area of the frame is 4x 2 + 34x.
You can see the graph in the embed window or the
photo last page.
When y(area) is 28, x is about 0.7 and -9.2.
So, the width of the frame is 0.7 because it cannot be minus.
DESCRIBE WHY/HOW A QUADRATIC MODEL CAN BE
HELPFUL IN THIS SITUATION
It’s difficult to calculate the width of the frame without the quadratic relation. When you
use the quadratic relations, you can determine the width even the area of the frame
change. Here is the photo of y = 23. When the area is 23, the width of the frame is
0.63. So, it will reduce your work when you use quadratic relations.
SITUATION 2
CREATE A QUESTION
THAT IS RELEVANT
TO THE SITUATION
3km/h
THE PROBLEM
There is a small car on the road.
The car goes to the destination and
goes back to the beginning.
3 hrs.
The speed of the wind is 2km/h.
The car takes 3 hours to go back to the beginning. the distance between the two places
are 15km.
Determine the speed of the car.
PROVIDE A SOLUTION TO YOUR QUESTION
x = the car ′ s speed on the road
y = the actual speed
When the car goes against the wind, y = x − 2.
On the other hand, y = x + 2 when the car downwind.
We can turn those speeds into times using:
time = distance / speed
total time = time against wind + time downwind = 3 hours
So,
15
15
Y = (𝑥−2) + (𝑥+2) = 3
First, multiplying through by (x − 2)(x + 2).
3(x − 2)(x + 2) = 15(x + 2) + 15(x − 2)
Expand this.
3(x − 4)2 = 15x + 30 + 15x − 30
Simplify.
3x 2 − 30x − 12 = 0
So, this is a quadratic relation.
Use the quadratic formula to
determine the x.
x=
[−b ± √(b2 − 4ac)]
2𝑎
x =
[−(−30) ± √((−30)2 − 4 × 3 × (−12))]
(2 × 3)
x =
[30 ± √(1044)]
6
x = −0.39/10.39
Car's Speed = 10.39 km/h
DESCRIBE HOW THE SITUATION REPRESENTS A
QUADRATIC RELATION
This problem has a little relationship to quadratic
Click the play button to see
relations. But you must use the quadratic relation
the graph
to solve the problem.
Because of time =
distance
.
𝑠𝑝𝑒𝑒𝑑
against the wind is
15
.
𝑥+2
15
.
𝑥−2
So, the time of
The time of downwind is
Then add them. Because the total time is 3
hours, so the equation can be
15
15
+ (𝑥+2)
(𝑥−2)
= 3.
Simplify the equation. It will be 3x 2 − 30x − 12 = 0.
The zeros are the final answer to this problem. The speed cannot be minus, so the
speed will be 10.39 km/h.
DESCRIBE WHY/HOW A QUADRATIC MODEL CAN BE
HELPFUL IN THIS SITUATION
You should better use the quadratic relation to solve this problem because you do not
know about how long did the one-way journey take. It’s easier to determine and your
mind can follow the question when solving.
When solve this problem, you must know time =
distance
𝑠𝑝𝑒𝑒𝑑
first. So, you can’t solve it by
only using quadratic relations. I think this problem is a little hard than the previous one.
Thanks to