Transcript Slide 1 - Forest Hills High School

Aim: How do we apply the quadratic equation?
Do Now: Given a equation:
y
x 6
x5
2
a) Find the coordinates of the turning point
b) 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.
Find the vertex of
yx 16
x
2
b 16 16
 8
x 
2a 2 2
 64
y
(8
)
16
(8
)64128
2
Use quadratic equation to find
the 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 Problems
Here is our picture:

Now we need to label the variables
Here are our variables:

y
x
z
We are trying to maximize the area, so we
write an expression for the area:
A = (x + z)y

y
x
z
The expression we have has too many
independent variables A = (x + z)y
so we need a relationship between the
variables x, y and z.
We know there
are 600 feet of
fence, so
2(x+z)+3y=600

y
x
z
We now use this relationship to eliminate
unwanted variables so we can write A as
an equation with one variable.
A  ( x  z) y
2( x  z)  3 y  600
2( x  z)  600  3 y
3
x  z  300  y
2

3 
A  300  y y

2 

3
3
2
A

(
300

y
)
y


y

300
y
2
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
32
A
 y 
300
y
2
is easily found: The x-coordinate is
b
y 
2a

300

2  (3/2)
 100
So, y =100. The other side of the
rectangle, x+z, is given by
3
x  z  300 y
2
3
 300 (100)
2
150

So, the dimensions are 100 by 150 and
the maximum area is their product,
15000.
