Dev

D.E.V. Project

Created by:

Tyler Fassezke

Problem One

• A landscaper has a job to lay a basketball court, the length of the court is 1.75X of the width. The man employing him will only pay $4500 and wants the court as large as possible. Lastly a square foot of cement cost $2.25 with a $300 initial cost. What is the largest he can make the court?

Step one, draw it out.

1.75X

1.75X

X X

Step two, find the amount of square feet possible.

1.75X

1.75X

X X1. $4500 = $2.25X + 300

-300 -300

2. $4200 = $2.25x

divide by 2.25

3. X = 1866.66 Square Feet.

Step 3, solve the equation.

• 1866.66 ft.²= X Multiplied by 1.75X

• Divide both sides by 1.75

• 1066.66 = X²

• √1066.66 = √X²

• 32.66 = X

Last Step! Insert X into the picture.

1.75X

1.75X

X X X = 32.66 Feet

57.15 Feet

57.15 Feet.

32.66 Feet.

32.66 Feet.

Problem Two

• Simplify the following:

• 5 + 3X+12 2X³+20x²-48X

(X-1) 2X-4 X² +9X+20

Step One, Start with the multiplication

3X+12 2X³+20x²-48X 2X-4 X² +9X+20

• Order of operations says multiply first.

• Next, factor the numerators and denominators.

Step Two, Factor.

3X+12 2X³+20x²-48X 2X-4 X² +9X+20

3(X+4) 2X(X²+10x²-24) 2(X-2) X² +9X+20

Factor out greatest common factor.

Factor again until you can do no more.

3(X+4) 2X(X+12)(X-2) 2(X-2) (X+5)(X+4)

Step Three, Simplify the multiplication.

3(X+4) 2X(X+12)(X-2) 2(X-2) (X+5)(X+4)

3(X+4) 2X(X+12) 2 (X+5)(X+4)

3 2X(X+12) 2 (X+5)

The (X+4) on the top and bottom cancel out.

The (X-2) on the top and bottom cancel out.

The 2 on the out side of the parenthesis on the top and bottom cancel out but the X stays.

Step Four, Combine the two fractions.

3 X(X+12) (X+5)

3X(X+12)

(X+5)

Step Five, Bring back the addition.

5 + 3X²+36X

X-1 (X+5)

Step Six, make denominator the same so that you can add the

fractions

(x+5) 5 + 3X²+36X (X-1)

(X+5) X-1 (X+5) (X-1)

Step Seven, multiply the nominator.

(X+5)5 + (3X²+36X)(X-1)

(X+5)(X-1) (X+5)(X-1)

F.O.I.L. to multiply the fractions.

5X+25 + 3X³-3X²+36X²-36x (X+5)(X-1) (X+5)(X-1)

5X+25 + 3X³+33X²-36X (X+5)(X-1) (X+5)(X-1)

Combine like terms.

Step Eight, Combine the fractions.

5X+25 + 3X³+33X²-36X

(X+5)(X-1) (X+5)(X-1)

3X³+33X²-31X+25

(X+5)(X-1)

Problem Three, Complete the Square.

5X²+15X+24=17

Step One, Subtract “C” from both sides.

5X²+15X+24=17

-24 -24

5X²+15X=-7

Step Two, factor out “A”

5X²+15X=-7

5(X²+5X__C__)=-7

Step Three, “C”=(b/2)²

5(X²+5X__C__)=-7

5/2=2.5

2.5²=6.25=C

5(X²+5X+6.25)=-7

Step Four, add the new “C” to both sides.

5(X²+5X+6.25)=-7

+5(6.25)

5(X²+5X+6.25)=24.25

Make sure to multiply “C” by the amount factored out. In this case Five.

Step Five, factor into binomial.

5(X²+5X+6.25)=24.25

5(X+2.5)²=24.25

It’s always

(X + or – 1/2B)²

Step Six, Divide by “A”

5(X+2.5)²=24.25

5

(X+2.5)²=4.85

Step Seven, Square root both sides.

√(X+2.5)²= √4.85

(X+2.5)=2.2

Step Eight, Get X alone and solve for X.

(X+2.5)=+or-2.2

-2.5 -2.5

X= -2.5+-2.2 or -2.5--2.2

X=-4.7,-.3

Graph the following function.

f(x)=-(X-6)(X+7)³(X-1)²(X+5)

Step One, Write down the X intercepts.

-(X-6)(X+7)³(X-1)²(X+5)

6 -7 1 -5

You are trying to find what X equals so you must take the opposite of the number that is in the parenthesis with it to equal zero.

Step Two, find where the “bounces” are.

-(X-6)(X+7)³(X-1)²(X+5)

6 -7 1 -5

• The graph bounces when there is a square. We have one at (X-1)². So we have one bounce.

Step Three, Calculate the number of Powers in the problem

-(X-6)(X+7)³(X-1)²(X+5)

1 + 3 + 2 + 1

There are 7 so it is odd therefore it will not start in the same direction as it ends.

Step Three, positive or negative?

The equation has a negative sign in the front, therefore the graph will start high

and end low.

Step Four, Draw the Graph.f(x)=-(X-6)(X+7)³(X-1)²(X+5)

Draw the dots and then connect them. Also the (X+7) is cubed so you must have a curve through. I aplogiese for the sloppyness but this is the best I could do in micro-soft powerpoint.