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William Low
Quadratic Function
MTH-4108-1
Unit 1
Determining the Maximum
Objective:
To solve problems, based on everyday situations, involving maximum yield, profit or
height. To fill in partially completed tables of values and find the value of x or y that
corresponds to the required maximum. Also to find the maximum value for situations
whose equation is given?
Mr. Pitt owns a peach orchard with 30 trees yielding an average of 400 peaches each. He wants to plant
more peach trees in order to get the maximum yield from his orchard. However, for each additional peach
tree planted , the average yield per tree will drop by 10 peaches. How many peach trees should Mr. Pitt
plant to ensure a maximum harvest?
Number of Peach
Trees to be Added
x
0
1
2
3
4
5
6
7
8
9
10
x
Total Number of
Peach Trees
Average Yield
per Tree
Total Yield of the Orchard
y
30
30 + 1
30 + 2
30 + 3
30 + 4
30 + 5
30 + 6
30 + 7
30 + 8
30 + 9
30 + 10
30 + x
400
400 – 10 ×1
400 – 10 ×2
400 – 10 ×3
400 – 10 ×4
400 – 10 ×5
400 – 10 ×6
400 – 10 ×7
400 – 10 ×8
400 – 10 ×9
400 – 10 ×10
400 – 10 × x
30 × 400= 12,000
(30 + 1) × (400 – 10 × 1) = 12,090
(30 + 2) × (400 – 10 × 2) = 12,160
(30 + 3) × (400 – 10 × 3) = 12,210
(30 + 4) × (400 – 10 × 4) = 12,240
(30 + 5) × (400 – 10 × 5) = 12,250
(30 + 6) × (400 – 10 × 6) = 12,240
(30 + 7) × (400 – 10 × 7) = 12,210
(30 + 8) × (400 – 10 × 8) = 12,160
(30 + 9) × (400 – 10 × 9) = 12,090
(30 + 10) × (400 – 10 × 10) = 12,000
(30 + x) × (400 – 10 × x) = y
Procedure for finding the value of x or y that corresponds to a maximum, using a table of values:
1. Read the problem carefully and define your variables.
2. Complete the table of values that represents the situation.
3. Find the ordered pair that corresponds to the maximum.
4. Depending on the question asked, find the value of x or y in the ordered pair that corresponds to
the required maximum.
Note: That the ordered pair corresponding to the maximum is the vertex of a parabola.
Procedure for finding the value of x or y that corresponds to a maximum, given an equation:
1. Determine what x and y represent.
2. Complete the table of values for the given values of x.
3. Find the ordered pair that corresponds to the maximum.
4. Depending on the question asked, find the value of x or y in the ordered pair that corresponds to
the required maximum.
C:/My Documents/My Work/Adult Ed/Math 436 Notes/Quadratic Function
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William Low
Unit 2
Equations Involving a Maximum
Objective:
To fill in a partially completed table of values and then find the equation that
represents the given situation.
See table in Unit 1 for the deduced equation.
Procedure for finding the equation for problems involving a maximum:
1. Read the problem carefully.
2. Fill in the partially completed table of values with at least three other values of x.
3. Complete the last line of the table by choosing x as the variable in the first column and fill in the
other columns as a function of x.
4. Do the multiplication indicated in the bottom right hand corner of the table and then write the
resulting expression as an equation of the form a x 2 + b x + c.
All quadratic equations can be written in the form:
y=ax2+bx+c
Where a, b, c   and a ≠ 0
Unit 3
Graphing an Equation of the Form y = a x 2
Objective:
To graph a second-degree equation of the form y = a x 2. The graph should include
the coordinates of the vertex, the axis of symmetry and its equation.
The fall of a pebble:
i.e.
1
at2,
u = 0 , a = 9.8 m / s 2
2
1
9 .8 m
4 .9 m 2 5 m
s= ×
× t2 =
× t = 2 × t2 [(3s)2 = 32 × s2 ]
2
2
2
s
s
s
s=ut+
Table of values for y = 5x2: Plot these ordered pairs on a Cartesian plane.
x
y = 5x2
y = − 5x 2
3
45
− 45
2
20
− 20
1
5
−5
0
0
0
−1
5
−5
−2
20
− 20
−3
45
− 45
 The graph of a quadratic function is always a parabola.
 The turning point of the curve is called the vertex. The vertex can represent either the maximum or
the minimum point of the parabola.
 The minimum point is the lowest point of the parabola.
 The maximum point is the highest point of the parabola.
 The two unlimited branches form the opening of the parabola. The parabola can open either upward
or downward.
 The parabola is divided into two symmetrical parts by a vertical line called the axis of symmetry.
Recall that the equation of a vertical line is x = k where k   .
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William Low

The parabola consists of an infinite number of points.
Tips on graphing a parabola:
1. For greater precision, choose at least two points to the left and two points to the right of the vertex.
2. We can choose any scale we want for the x-axis and the y-axis. For example, to draw the parabola
y = x 2, we could use a scale of 0.5 cm = 1 unit for the x-axis and the y-axis. If the equation is
y = 10 x 2 , a scale of 0.5 cm = 1 unit would be acceptable for the x-axis, and we could choose a
scale of 0.5 cm = 5 units or a scale of 0.5 cm = 10 units for the y-axis.
3. As far as possible, it is preferable to keep using the scale 0.5 cm = 1 unit for both axes in order to
graph the parabola as accurately as possible.
What would happen to our parabola y = 5x2 if y = − 5x2 ? See table above and plot.
If a is positive (a > 0) , the parabola opens upward.
If a is negative (a < 0, the parabola opens downward.
The greater the value of | a |, the narrower the parabola.
The smaller the value of | a |, the wider the parabola.
The absolute value | a | of a number is the value of the number regardless of its sign.
To sum up, a plays two roles:
1. The sign of a ( + or − ) determines in which direction the parabola opens :
If a > 0, the parabola opens upward.
If a< 0, the parabola opens downward.
2. The relative size of a in terms of the absolute value determines the width of the parabola :
If | a | > 1, the parabola is narrow
If | a | < 1, the parabola is wide.
Unit 4
Graphing an Equation of the Form y = a x 2 + c
Objective :
To graph equations of the form y = a x 2 + c , where a is a rational number between
− 5 and 5 (a ≠ 0) and where c is a rational number. The graph should include the
coordinates of the vertex, the axis of symmetry and its equation.
If you let a ball fall from the top of the Olympic Stadium’s tower, which is 190 m above the ground, the
height (h) of the ball after t seconds is represented by the equation :
h = − 16 t 2 + 190
(air resistance not considered)
When h = 0, t = 3.45 seconds i.e. it would take the ball 3.45 s to reach the ground.
This is an example of an equation of the form y = a x 2 + c.
x
−2
−1
0
1
2
y=x2
4
1
0
1
4
y=x2+3
7
4
3
4
7
y=x2−3
1
−2
−3
−2
1
y=−x2
−4
−1
0
−1
−4
C:/My Documents/My Work/Adult Ed/Math 436 Notes/Quadratic Function
y=−x2+3
−1
2
3
2
−1
y=−x2−3
−7
−4
−3
−4
−7
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William Low
Draw these parabolas and discuss their relevant points.
e.g. opening , width , axis of symmetry , abscissa , ordinate.
A translation is the movement of an object such that every point on that object is moved the same
distance and in the same direction in the plane.
Investigate the parabola where the value of a in the equation y = a x 2 + c is negative. See the above table
of values.
Note the similarities between the parabolas:
 They all open downwards.
 They all have the same width.
 They all have the same axis of symmetry (x = 0).
 The all have the same abscissa at the vertex but they do not have the same ordinate at the vertex.
If c is positive (c > 0), the curve moves above the x – axis.
If c is negative (c < 0), the curve moves below the x – axis.
For all equations of the form y = a x 2 + c , the equation of the axis of symmetry will be x = 0.
In addition, for any equation of the form y = a x 2 + c , the coordinates of the vertex are (0 , c ).
If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward.
Unit 5
Solving a Second-Degree Equation by Factoring
Objective:
To find the zeros of an equation, describe the steps involved in solving a problem and
verify these steps by using the factoring method.
Solving a quadratic equation involves finding the values of x when y = 0. These values of x are called the
zeros or the roots of the equation.
The following two algebraic methods can be used to solve this type of second-degree equation.
1. Factoring.
2. The quadratic formula.
Summary of the five factoring methods:
1.
2.
3.
4.
5.
Factoring by removing the common factor.
Factoring by grouping.
Factoring trinomials of the form x2 + b x + c or
Factoring trinomials of the form ax2 + b x + c or
Factoring a difference of squares.
x2 + b x y + cy2.
ax2 + b x y + cy2.
If the product of two numbers is zero, then one of these numbers is equal to zero.
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William Low
Procedure for solving a quadratic equation:
1. Write the quadratic equation in the form y = ax2 + b x + c.
2. Factor the polynomial.
3. Consider the two possibilities and find the values of x.
4. Check the results against the initial equation.
5. Give the values of x.
Unit 6
Solving a Second-Degree Equation Using the Quadratic Formula
Objective:
To use the quadratic formula to find the zeros of a second-degree equation.
Quadratic Formula:
x=
 b  b 2  4ac
2a
Procedure for solving a quadratic equation using the quadratic formula:
1. Write the quadratic equation in the form y = ax2 + b x + c.
2. Find the values of a, b, and c.
3. Replace a , b , and c with these values in the quadratic formula :
 b  b 2  4ac
2a
4. Calculate and simplify, if possible.
5. Check your answers against the original equation.
6. Give the values of x.
x=
The discriminant = Delta (Δ) = b 2 − 4 a c
1. If Δ is positive (Δ > 0), the equation has two solutions.
2. If Δ is null (Δ = 0), the equation has one solutions.
3. If Δ is negative (Δ < 0), the equation has no solution.
Unit 7
Graphing a Second-Degree Equation
Objective:
To graph second-degree equations and identify the vertex, the axis of symmetry, the
y-intercept, and , if possible , the zeros.
The abscissa of the vertex is given by the formula:
The ordinate of the vertex is given by the formula:
b
2a

y=
4a
x=
where Δ = b 2 − 4 ac
The coordinates of the vertex of a parabola are:
b 
,


 2a 4a 
C:/My Documents/My Work/Adult Ed/Math 436 Notes/Quadratic Function
where Δ = b 2 − 4 ac
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William Low
Procedure for finding the vertex of a parabola:
1. Write the quadratic equation in the form y = ax2 + b x + c.
2. Find the values of a, b, and c.
3. Find the discriminant using the formula Δ = b 2 − 4 a c.
b 
4. Find the coordinates of the vertex using the formula 
,
.
 2a 4a 
5. Check whether the point obtained in Step 4 is correct by substituting its coordinates for the
variables x and y in the original equation.
In all quadratic equations, the y-intercept is represented by the point (o, c).
Procedure for sketching the parabola whose equation is given:
1. Write the quadratic equation in the form y = ax2 + b x + c.
2. Identify the values of a, b, and c and indicate whether the parabola opens upward or downward
according to the sign of a.
3. Find the discriminant using the formula Δ = b 2 − 4 a c.
4. Indicate the number of zeros or roots.
5. Find the zeros, if any, by applying the factoring method or the quadratic formula. Find the
coordinates of these zeros and plot the corresponding points in the Cartesian plane after choosing
the appropriate scale.
b 
,
6. Find the coordinates of the vertex using the formula 
.
 2a 4a 
b
b
7. Find the equation of the axis of symmetry x =
(
being the abscissa of the vertex) and
2a
2a
draw it as a dotted line in the Cartesian plane.
8. Identify the point representing the y-intercept (0 , c) and plot it in the Cartesian plane.
9. In the Cartesian plane, plot the point symmetric to the y-intercept with respect to the axis of
symmetry.
10. Draw up a table of values if the points found previously are not sufficient to obtain a complete and
accurate parabola.
11. Connect all the resulting points in a continuous curve to obtain the sketch of the parabola.
Unit 8
Determining the Maximum or Minimum Given a Second-Degree Equation
Objective:
To find the value of x and of y at the maximum or minimum point of a parabola and
to show the steps in the solution. Each problem relates to business or the physical
sciences and deals with a situation that represents a quadratic function whose
equation is given.
Procedure for finding the maximum or the minimum point using a quadratic equation:
1. Read the problem to determine what x and y represent.
2. Find the values of x and y at the maximum or minimum point i.e. at the vertex of the
b 
,
parabola 
.
 2a 4a 
3. Give the value of x or y depending on the question.
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William Low
Unit 9 Solving a Problem that can be written as a Second-Degree equation
Objective:
To solve problems by expressing them as second-degree equations, describing the
steps in the solution and giving the required values.
Problem solving procedure:
1. Define the problem, i.e. identify what you are looking for and express it as a variable.
2. Write the equation by translating the word problem into mathematical language.
3. Convert the equation to the form ax2 + b x + c = 0.
4. Solve the equation by using the most appropriate method (factoring or the quadratic formula).
5. Determine the required value(s) according to the information in Step 1and disregard any irrelevant
values.
6. Check the result against the initial word problem.
Consecutive numbers are numbers that follow one after the other. The difference between each pair of
consecutive numbers is 1. Thus, 3 and 4 are consecutive numbers as are − 2 and − 1.
Areas:
Square = s2; Rectangle and a Parallelogram = b × h;
D  d 
Triangle =
b  h  ;
2
where “D” is the long diagonal and “d” the short diagonal.
2
B  b  h where “B” is the long base and “b” is the short base.
Trapezoid =
2
Rhombus =
Final Review
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