Class notes from 8/29

Try same ideas with

3x + 7 y = 9

81 x – 35 y = 17

and with

2 x + 32 y = 7

3 x – 5 y = 8.

Next: get model by simul eq, compare, etc..

The form of a quadratic function is

• y = a t^2 + b t + c.

If we use the data points (10,80), (46, 50) and (99,20) we can get three simultaneous linear equations.

(10, 80) means y = 80 when t = 10. In this case our equation will be

• 80 = a ( 10^2) + b ( 10) + c.

The point (46, 50) gives us the equation

• 50 = a (46^2) + b(46) + c.

The point (99, 20) gives the equation

• 20 = a(99^2) + b(99) + c.

Squaring and rearranging we see that we end up with the equations

100 a + 10 b + c = 80

2116 a + 46 b + c = 50

9801 a + 99 b + c = 20.

How we gonna solve this system?

Eliminate one of the variables to get 2 equations in 2 unknowns. Pick on c:

Now we have the system

-2016 a – 36 b = 30

–9701 a – 89 b = 60.

To solve eliminate b. Multiply 1^st by 89 and 2d by –36 to get

89 [ -2016 a – 36 b = 30 ]

-36 [ –9701 a – 89 b = 60 ]

so

-179,424 a – 3204 b = 2670

349,232 a + 3204 b = -2160.

Sum is

169,812 a = 510 so

a = 510 / 169,812 = .003003 (approx.).

Now plug a into -2016 a – 36 b = 30 to get

-2016 (.003003) – 36 b = 30

-6.054 - 36 b = 30

-36 b = 36.054

b = -1.0015.

Now plug a and b into the first equation:

100 a + 10 b + c = 80 so

100 ( .003003) + 10 (-1.0015) + c = 80

.3003 – 10.015 + c = 80

-9.712 + c = 80

c = 89.71.

So our y = a t^2 + b t + c model is now

• y = .003003 t^2 – 1.015 t + 89.71.

The Excel solution, which used ALL the data points and not just the three we chose, was

• y = .0031 t^2 –1.0089 t + 89.776,

so we're pretty close.

According to the rounded-off model y = .003 t^2 – t + 90:

• Plug in t = 60 to get y = .003 ( 60^2) – 60 + 90 = 40.8.

• Occurs when y = 25. Substituting we get: