Aim: How do we apply the quadratic equation?

Do Now: Given a equation: 562 xxy

a) Find the coordinates of the turning pointb) If y = 0, find the values of x

HW: Worksheet

The turning point is (3,4). This is the maximum point.

Notice that the equation has a negative leading coefficient, then there is a maximum point.

To find the values of x when y = 0, we simply replace y by 0 then solve the equation for x.

A quadratic equation and the parabola can be applied in many real life situations.

Here is the simple example:

We can treat the equation as the parabola of the advancing path of a baseball. The maximum point the where the ball reaches its maximum height.

The x can be use as the number of seconds and the y can be the height in meter or feet.

When y = 0 the x are 1 and 5. That means when time is 0 second the height is 0 meter or feet, when time is 5 seconds the ball comes back to the ground.

xxy 162 Find the vertex of

2

16

2

a

bx 8

2

16

)8(16)8( 2 y 12864 64

Use quadratic equation to find Use quadratic equation to find the maximum or minimumthe maximum or minimum

Problem:A rectangular playground is to be fenced

off and divided into two by another fence parallel to one side of the playground. Six hundred feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area?

We need to draw a rectangular field divided in two pieces by a fence parallel to one side.

More Joy of Word ProblemsMore Joy of Word Problems

Here is our picture:

Here are our variables:

y

x z

Now we need to label the variables

A = (x + z)y

y

x z

We are trying to maximize the area, so we write an expression for the area:

The expression we have has too many independent variables

so we need a relationship between the variables x, y and z.

A = (x + z)y

y

x z

We know there are 600 feet of fence, so

2(x+z)+3y=600

We now use this relationship to eliminate unwanted variables so we can write A as an equation with one variable.

A(x z)y2(x z) 3y 600

2(x z) 600 3y

x z300 3

2y

A 300 3

2y

y

yyyyA 3002

3)

2

3300( 2

We notice that this is a quadratic equation. Since the leading coefficient is negative, its graph is a parabola which opens downward. Therefore the vertex of this parabola is the place where the maximum occurs.

The vertex of the parabola

is easily found: The x-coordinate is

y b2a

300

2 ( 3/2)

100

yyA 3002

3 2

So, y =100. The other side of the rectangle, x+z, is given by

So, the dimensions are 100 by 150 and the maximum area is their product, 15000.

x z300 3

2y

300 3

2(100)

150