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Kerr
MPM2D
Quadratic Equations
6.6 Solving Problems Using Quadratic Models
The diagram below summarizes the strategies used to solve quadratic
word problems. It is important to read the question carefully so that
you choose the correct strategy.
Quadratic
Word
Problems
Optimum Value
(Maximum or
Minimum)
Complete the
Square
Solve for the axis of
symmetry and
substitute back into
the original equation
The value of x
Factor
Use the Quadratic
Formula
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MPM2D
Quadratic Equations
Examples:
1.
An object is thrown from the top of a building. Its path is
modelled by h = -5t2 + 8t + 10, where h is the height in m, and t is
the time in seconds.
a)
When will the object hit the ground? Round your answer to the
nearest hundredth.
Since the question said WHEN,
we are solving for time. When
the object hits the ground, the
height is equal to zero.
Substitute zero for the height
in the equation and solve using
the quadratic formula. Notice
the question said nothing about
a maximum or minimum so we
will not be completing the
square or finding the axis of
symmetry.
Therefore, the object hits the ground after __________
seconds.
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MPM2D
Quadratic Equations
b)
When is the object at a height of 12m? Round your answer to
the nearest hundredth.
This question is very similar to
part a. Substitute twelve for
the height into the equation
and solve using the quadratic
formula.
Therefore, the object is at a height of 12m after __________
seconds and __________ seconds.
c)
What is the maximum height of the object and when does it
occur? Round your answer to the nearest hundredth.
We can solve this problem two different ways.
Method #1
(from part a)
_______________
_______________
(from part b)
_______________
_______________
The axis of symmetry is __________.
In part a and part b,
we found two points
with the same ycoordinate. These
points can be used to
find the axis of
symmetry by adding
them together and
dividing by two. We
should be able to get
the same answer with
both sets of points.
Kerr
MPM2D
Quadratic Equations
x=
b
2a
h = -5t2 + 8t + 10
We can check to see if this is
correct by using the shortcut!
Substitute the axis of
symmetry back into the
original equation to solve
for the maximum height.
Therefore, the maximum height of the object is __________ at
__________ seconds.
Method #2
h = -5t2 + 8t + 10
Since we need the
maximum height,
complete the square.
Therefore, the maximum height of the object is __________ at
__________ seconds.
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MPM2D
Quadratic Equations
2.
Find the dimensions of a rectangle that has a perimeter of 60
feet and the greatest possible area.
Perimeter of a rectangle is 2l + 2w. The
question said the perimeter is 60. Substitute 60
into the formula for perimeter and isolate w.
We want to maximize area. Area of
a rectangle is length x width.
Substitute w = 60 - l into the
equation so we only have one
unknown.
We must maximize the equation A = l(30 - l), where length (l) is
the independent variable (x) and A (area) is the dependent
variable (y).
We can solve this problem two different ways.
Method #1
Since the expression is already
factored, let's solve for the zeros,
then find the axis of symmetry. We
can substitute the axis of symmetry
back into the original equation to
find the maximum value.
A = l(30 - l)
axis of symmetry
solve for w
Maximum Value
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MPM2D
Quadratic Equations
Therefore, the greatest possible area is __________ feet2 when
length is _____ feet and width is _____ feet.
Method #2
A = l(30 - l)
Expand and simplify to convert the
relation to standard form. Then,
complete the square since we need
the maximum value.
Substitute _____ into ____________ to solve for the width.
Therefore, the greatest possible area is __________ feet2 when
length is _____ feet and width is _____ feet.
Homework: Page 358 #3-8, 12, 13, 14, 17