September 20, 2006

Word Problems Involving Parabolas

Hi everyone.
Well, we started today's class by solving two problems which were due as home work. We (most of the class) had a hard time trying to complete the square because of one of the questions because of FRACTIONS.
Mr. K. used another problem (easy one) to illustrate the use of fractions.
The easy one was: y=3x^2 + 12x +4
to complete the square we can factor the 3 out.
y = 3[x^2+4x+4] then complete the square
y = 3(x+2)^2

The other way was to use fractions.
y = 3x^2 + 4x +4 complete the square while placing the "+4" outside the bracket.
y = 3[ x^2 +4 +_ - _ ] +4 now lets try to complete the square inside the bracket


Remember:
In order to complete the square in the general formula (AX^2 + BX + C) ,C have to be (B/2)^2 in other words (B*1/2)^2
Therefore to find the C for y=3[x^2 + 4x +__ -__] +4 we have to multiply by 1/2 then square. (4*1/2)

Therefore
y = 3[x^2 +4 +(4*1/2)^2 - (4*1/2)^2] +4
y = 3[x^2 +4 +(4/2)^2 -(4/2)^2] +4
{{{Reminder: don't square the +ve number we added, because we will soon need its square root.}}}
y = 3[x^2 +4 +(4/2)^2 -(16/4)] +4
y = 3[(x+4/2)^2 - 16/4] +4 ... MULTIPLY THE 3 IN.
y = 3(x+4/2) ^2 - 48/4 + 4 ...to add these two numbers we have got to have the same denominator.
Therefore
y = 3(x+4/2)^2 - 48/4 + 48/4
y =3(x+4/2)^2
y = 3(x+2)^2

"NOW LETS DO THE REAL FRACTION STUFF"

Lets change this (from general form to the standard form):
y = 6 + 5x - 2x^2 ...rearrange
y = -2x^2 +5x +6...factor -2 out
y = -2[x^2 + 5/2x +__-__] +6...to get the absolute value which completes the square we have to get the middle term times 1/2 then square it. = (5/2 * 1/2)
y = -2[x^2 +5/2x +(5/2 * 1/2)^2 -(5/2 * 1/2)^2] +6

y = -2[x^2 + 5/2x + (5/4)^2 -(5/4)^2] +6...remember: don't square (5/4)^2 because we will soon need its square root.
y = -2[(x+5/4)^2 - 25/16] +6...Multiply the -2 in.
y = -2(x+5/4)^2 + 50/16 +6...Simplify the fraction.
y = -2(x+5/4)^2 + 25/8 +6 ... change 6 into a fraction where by its denominator is 8
y = -2(x+5/4)^2 +25/8 + 48/8
y = -2(x+5/4)^2 +73/8
Now we should be able to estimate these fractions. For example 73/8 is about 9.1 & -5/4 is about -1.3. The vertex is (-5/4,73/8)...or...(-1.3,9.1)


WORD PROBLEMS INVOLVING PARABOLAS

Usually there are 3 types of these problems. These problems wont be given in the standards form. They will either be in the general/factored form.


ONE :
Grad tickets are sold for $20 each, and 300 students are willing to buy them at that price. For every $5 increase in price there are 30 fewer students willing to buy tickets what is the maximum Revenue?
Rephrase the Q: how many times should they increase the price in order to get the maximum Revenue ($$$$$$$ get it?)

Legend:
X= increase in price by $5
Therefore:
Revenue= (cost/ticket) (number of students willing to attend)

Revenue as a function of X (depends on X)
This is in the factored form

(x) How many times the price is raised by $5

REVENUE (X) = (20+5x) (300-30x)

(300-30x) The original students willing to attend minus the number of students who wont be willing to buy tickets (3o students) each times x raises by one.

(20 + 5x) The original price + $5 each time value of X increases by one.

R (x) = ($20 + $5x) (300 - 30x)
Roots (Zeros) = -4 and 10
because (20 + 5(-4)=0 and (300 -30(10))=0

The axis of symmetry or X= -4 +10/2 = 3

Therefore R(3)= [$20+$5(3)] * [300-30(3)]
R(3)= ($35/ticket) (210 students will attend)
Revenue(3)= $7350

TWO:
A toy Rocket is shot upwards from a height of 6 meters. its height, in meters, above the ground after "t" seconds is given by h(t)=6 + 80t -5t^2 ...
h(t)= -5t^2 + 80t +6 <<<<<
a} After how many seconds does it reach its maximum height
?
Rephrase: what is the axis of symmetry?

h(t)= -5t^2 + 80t +6
h(t)= -5[ t^2 -16 +_ -_] +6
h(t)= -5[ t^2 -16 +64 - 64] +6
h(t)= -5[ (t-8)^2 -64] +6
h(t)= -5(t-8)^2 + 320 +6
h(t)= -5(t-8)^2 +326

Since h in the above a(x-h)^2+k equals 8. Therefore the axis of symmetry= 8
Rephrase: The rocket will reach its max. point after 8 seconds.

b} What is its max. height?
Rephrase: What is the h co-ordinate of the vertex, or what is the max point of the parabola? (because it opens downwards ...).

Since the vertex = (8,326), the max point, or the h co-ordinate of the vertex= 326 meters

I will insert the graph tomorrow, and if there is something wrong with calculation i will make sure i check with Mr. K
THIRD:
Find the maximum rectangular area that can be enclosed by 160 of fencing if one side of the rectangle is an existing wall.


Width*2 + Length (one side)= 160
Therefore: Length = 160-2w

Lets find the table of values.
Width ....Length ....Area
10 ........140 .......1400
20 ...........120 ......2400
30....... 100....... 3000
40 ...........80...... 3200
50 ...........60..... 3000
60 ..........40........ 2400
and decreases

Area(w)= (160 - 2w)w
Roots
0=160 - 2w
160 = 2w
80=w

Therefore the roots are zero and 80. Zero because if we replace w by zero it gives us zero
Area(0)= (160-2(0))0
= (160)(0)
=ZERO!

h=0+80/2
Therefore h=40!


Uffffff...let me take a breath...ok...my last thing. THE NEXT SCRIBE WILL BE.
\You are right Lauressa (: ...You are the one!



3 comments:

  1. the guy with glasses9/22/2006 10:36 AM

    studing for my quadratic funtion unit text today. so i thought i come to check out mr.k blog for help.
    it just get me thinking how bad my blogs is compare yours nataneal hahah. good job. but your guys are kind of behind? i think.
    anyways. good job. keep up the good work.

    ReplyDelete
  2. i mean test. hahaha not text

    ReplyDelete
  3. This comment has been removed by the author.

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