1. Chapter 3
Quadratic Functions

2. 3.1 – Investigating Quadratic functions in vertex form
Chapter 3:
Quadratic Functions

3. handout
Work through the Activity Sheet which outlines the investigation described on page Make sure to answer all the questions to your fullest ability since this a summative assessment. Feel free to use the Y= function on your calculator to help you graph the functions.

4. definitions
A quadratic function is a function f whose value f(x) at x is given by a polynomial of degree two.
Example: f(x) = x2, the simplest form of a quadratic function
A parabola is the symmetrical curve of the graph of a quadratic function.
A vertex (of a parabola) is the lowest point of the graph (if the graph opens upward) of the highest point of the graph (if the graph opens downward).

5. More terms
The minimum value (of a function) is the least value in the range of a function.
The maximum value (of a function) is the greatest value in the range of a function.
The axis of symmetry is a line through the vertex that divides the graph of a quadratic function into two congruent halves. It is defined by the x-coordinate of the vertex.

6. Vertex form f(x) = a(x – p)2 + q What does the parameter a affect?
Parameter a determine the orientation and shape of the parabola.
The graph opens upward if a is positive, and downward if a is negative.
If -1 < a < 1, the parabola is wider compared to the graph f(x) = x2.
If a > 1 or a < -1, the parabola is narrower compared to the graph f(x) = x2.

7. Vertex form f(x) = a(x – p)2 + q
What does the parameter q affect?
Parameter q translates the parabola vertically q units relative to the graph of f(x) = x2.
The y-coordinate of the parabola’s vertex is q.

8. Vertex form f(x) = a(x – p)2 + q What does the parameter p affect?
Parameter p translates the parabola horizontally p units relative to the graph of f(x) = x2.
The x-coordinate of the parabola’s vertex is p.
The equation of the axis of symmetry is x – p = 0 or x = p.

9. example f(x) = a(x – p)2 + q
Determine the following characteristics for y = 2(x + 1)2 – 3.
The vertex
The domain and range
The direction of the opening
The equation of the axis of symmetry.
Then, sketch each graph.
Remember:
f(x) = a(x – p)2 + q
By transformations:
p = -1, q = -3, so the vertex is (-1, -3)
Since q = -3, the range is {y|y ≥ -3, y R}
The domain is {x|x R}
a = 2, is positive, so it opens upward.
The axis of symmetry happens at the x-value of the vertex, so at x = -1.

10. example f(x) = a(x – p)2 + q a = –2, p = 0, q = 3  f(x) = –2x2 +3
Determine a quadratic function in vertex form for the graph.
Remember:
f(x) = a(x – p)2 + q
Now we can pick any point that we know on the graph, and substitute the values into the equation:
We see (1,1):
1 = a(1)2 + 3
a = –2
Substitute what we know:
f(x) = a(x – 0)2 + 3
f(x) = ax2 + 3
The vertex is at (0, 3).
p = 0
q = 3
The graph opens downwards, so a will be negative.
a = –2, p = 0, q = 3
 f(x) = –2x2 +3

11. Try it!

12. example
Determine the number of x-intercepts for each quadratic function.
a) f(x) = 0.8x2 – 3 b) f(x) = 2(x – 1)2 c) f(x) = –3(x + 2)2 – 1
What will affect the number of x-intercepts?
The direction of the opening of the parabola
The location of the vertex a)
Opens upward
Vertex at (0, -3) b)
Opens upward
Vertex at (1, 0) c)
Opens downward
Vertex at (-2, -1)
0 x-intercept
2 x-intercepts
1 x-intercept

13. example
The deck of the Lions’ Gate Bridge in Vancouver is suspended from two main cables attached to the tops of two supporting towers. Between the towers, the main cables take the shape of a parabola as they support the weight of the deck. The towers are 111 m tall relative to the water’s surface and are 472 m apart. The lowest point of the cables is approximately 67 m above the water’s surface.
Model the shape of the cables with a quadratic function in vertex form.
Determine the height above the surface of the water of a point on the cables that is 90 m horizontally from one of the towers. Express your answer to the nearest tenth of a metre.
a) The vertex will be at the lowest point on the cables, so at 67 m above the water’s surface. Consider the vertex to be the origin, or (0,0). The highest points are 111 m above the water’s surface, so they have a value of y = 111 – 67 = 44, and they are 472 m apart, so they have x-values of x = 472/2 = 236 and x = —236.

14. Example continued So, a = 11/13924, p = 0 and q = 0
The vertex is at (0,0)
p = 0
q = 0
f(x) = ax2
Substitute (236, 44):
44 = a(236)2
a = 44/2362
a = 11/13924
So, a = 11/13924, p = 0 and q = 0
 f(x) = (11/13924)x2

15. Example continued 16.839 + 67 = 83.8 m above the water’s surface.
b) Determine the height above the surface of the water of a point on the cables that is 90 m horizontally from one of the towers. Express your answer to the nearest tenth of a metre.
Recall: f(x) = (11/13924)x2
It is m above the low point (which is 67 m above the water’s surface), so we need to add the two numbers:
90 m away from one of the towers, so x = 236 – 90 = 146.
f(x) = (11/13924)1462
f(x) = …
= 83.8 m above the water’s surface.

16. P # 3, 7, 8, 10, 12, 15, 17, 20, 21, 22
Independent Practice

17. 3.2 – Investigating quadratic functions in standard form
Chapter 3:
Quadratic Functions

18. Standard form Get out your calculators!
The form f(x) = ax2 + bx + c, or y = ax2 + bx + c, where a, b, and c are real numbers are a ≠ 0, is considered standard form.
Get out your calculators!
Try graphing the function f(x) = –x2 + 4x + 5
What’s the graph look like? What’s its symmetry?
Does it have a maximum y-value?
What about a minimum y-value?

19. The effect of parameter c
Graph these functions
f(x) = –x2 + 4x + 10
f(x) = –x2 + 4x
f(x) = –x2 + 4x - 5
What do you think the parameter c affects?

20. The effects of parameter A
Graph these functions
f(x) = –4x2 + 4x + 5
f(x) = –2x2 + 4x + 5
f(x) = x2 + 4x + 5
f(x) = 2x2 + 4x + 5
What do you think the parameter a affects?

21. The effects of parameter B
Graph these functions
f(x) = –x2 + 2x + 5
f(x) = –2x2 + 5
f(x) = –x2 – 2x + 5
f(x) = –x2 – 4x + 5
What do you think the parameter b affects?

22. In summary b = –2ap c = ap2 + q p = –b/(2a) q = c – ap2
For the standard form, f(x) = ax2 + bx + c:
a determines the shape and whether the graph opens upward (positive) or downward (negative)
b influences the position of the graph
c determines the y-intercept of the graph
Some equations that relate a, b, and c to p and q:
b = –2ap
c = ap2 + q
p = –b/(2a)
q = c – ap2

23. example Opens upward y-intercept at (0,0), and has a value of 0
For each graph of a quadratic function, identify:
the direction of the opening
the coordinates of the vertex
the maximum/minimum value
the equation of the axis of symmetry
the x-intercepts and the y-intercept
the domain and range
On this graph, we can just see all of the things that we’re looking for!
Opens upward
Vertex at (1, -1)
Minimum at y = -1, where x = 1.
the equation of the axis of symmetry is x = 1
y-intercept at (0,0), and has a value of 0
x-intercept at (0,0) and (2,0), and have values of 0 and 2
D : {x | x E R}
R : {y ≥ 1 | y E R}

24. example For each graph of a quadratic function, identify:
the direction of the opening
the coordinates of the vertex
the maximum/minimum value
the equation of the axis of symmetry
the x-intercepts and the y-intercept
the domain and range
Opens upwards
Vertex:
We can see that it is at x=3, so we can put that into the equation to find the y-value.
f(x) = 2(3)2 – 12(3) +25
f(x) = 18 – = 7
So the vertex is (3, 7)
The minimum is the vertex, so it’s at y = 7 where x = 3.
The axis of symmetry is x = 3.
There are no x-intercepts.
The y-intercept is at y = 25 (we can check this by substituting x = 0 into the equation).
D : {x | x E R}
R: {y ≥ 7 | y E R}

25. example
A frog sitting on a rock jumps into a pond. The height, h, in centimetres, of the frog above the surface of the water as a function of time, t, in seconds, since it jumped can be modeled by the function h(t) = –490t t Where appropriate, answer the following questions to the nearest tenth.
Graph the function.
What is the y-intercept? What does it represent in this situation?
What maximum height does the frog reach? When does it reach that height?
When does the frog hit the surface of the water?
What are the domain and range in this situation?
How high the frog 0.25 s after it jumps? a)
You will need to adjust your windows if you want to graph this on your calculator.

26. Example (continued)
On your calculator in the graph window, hit TRACE, it will give you the y-intercept in the Y = value.
y-intercept is h = 25.
The y-intercept represents the frog at the start, when time = 0, so it’s how high it was sitting to begin with.
D: {t |t E R}
R: {h ≤ 36.5 |h E R}
You can find the maximum by pressing 2ND + TRACE + 4, then hitting enter when the target is to the left of the max, then move it over to the right and hit enter twice more.
It’s at h = 36.5 cm high
At t = 0.15 seconds
The frog hits the water at the x-intercept, which you can find by 2ND + TRACE + 2, then hitting enter at the left bound and twice at the right bound.
t = seconds
After 0.25 s, you can hit TRACE , you see that h = 31.9 cm.

27. example
A rancher has 100 m of fencing available to build a rectangular corral.
Write a quadratic function in standard form to represent the area of the corral.
What are the coordinates of the vertex? What does the vertex represent in this situation?
Sketch the graph for the function you determined in part a).
Determine the domain and range.
Identify any assumptions you made in modeling this situation mathematically.
Let L represent the length and W represent the width, and let A be the area.
A = LW
P = 2L + 2W = 100
L + W = 50
L = 50 – W
A = (50 – W)W
 A = 50W – W2
b) Recall that p = –b/2a
p = –50/2(-1)
p = 25
Now, we can simply substitute W = 25 into the equation:
A = 50(25) – 252 = 625
Vertex is (25, 625)

28. Example continued d) D: {W | W E R} R: {A ≤ 625| A E R}
e) What are some assumptions we made?
That the rancher will use all of the fencing.
We also assume that any length from 0 to 50m would be appropriate for the width or length.
Would a 1 m wide rectangle be reasonable?
What about the shape of his field?
d) D: {W | W E R}
R: {A ≤ 625| A E R}

29. Pg , # 5, 6, 8, 9, 11, 13, 15, 18, 21-24
Independent Practice

30. 3.3 – Completing the square
Chapter 3: Quadratic Functions

31. handout
Completing the Square is a method used to convert quadratic functions from standard to vertex form.
Why might we want to be able to convert into vertex form?
Complete the activity described in the handout and answer all the questions to the best of your ability. You will need algebra tiles.

32. Completing the square
Completing the square involves adding a value to and subtracting a value from a quadratic polynomial so that it contains a perfect square trinomial. You can then rewrite this trinomial as the square of a binomial (in other words, rewrite it in vertex form).
Step 1: Put brackets around the x2- and the x-term.
Example:
y = x2 – 8x + 5
y = (x2 – 8x) + 5
y = (x2 – 8x + 16 – 16) + 5
y = (x2 – 8x + 16) – 16 +5
y = (x – 4)2 –
y = (x – 4)2 – 11
Step 2: Add/Subtract (b/2)2 inside of the brackets.
Step 3: Pull the term you don’t need outside of the brackets.
Step 4: Factor what’s in the brackets.
When will the term you add inside the brackets need to be positive? When will it be negative?
Step 5: Simplify!

33. Completing the square y = x2 – 8x +5  y = (x – 4)2 – 11
Now what do we know about the graph of this function?
y = x2 – 8x +5
 y = (x – 4)2 – 11
Now we know that the vertex is (4, –11).
We always knew it would open upwards
Make a table of values: x y 5 2 -7 4
-11 6 8

34. EXAMPLE Rewrite f(x) = 3x2 – 12x – 9 in vertex form.
First: simplify the expression by seeing if there are any terms that we can factor out.
f(x) = 3(x2 – 4x) – 9
f(x) = 3(x2 – 4x + 4 – 4) – 9
f(x) = 3[(x2 – 4x + 4) – 4] – 9
f(x) = 3[(x – 2)2 – 4] – 9
f(x) = 3(x – 2)2 – 12 – 9
f(x) = 3(x – 2)2 – 21
What number can we factor out of the first two terms?
What number do we need to add/subtract inside the brackets?
Why do I need an extra set of brackets in the third line?
Now, what would the vertex of the function be?

35. example Convert the function y = 4x2 – 28x – 23 to vertex form.
Verify that the two forms are equivalent.
y = 4x2 – 28x – 23
What can we factor out of the first two terms?
y = 4(x2 – 7x) – 23
We can use either fractions or decimals:
y = 4(x2 – 7x – 3.52) – 23
y = 4[(x2 – 7x ) – 12.25] – 23
y = 4[(x – 3.5)2 – 12.25] – 23
y = 4(x – 3.5)2 – 4(12.25) – 23
y = 4(x – 3.5)2 – 72
What is an easy way to verify that these two functions are actually the same?
Try entering them both into your Y= screen on your calculator, and check that the graphs line up!

36. pG , # 3, 4, 8, 11, 12, 14, 15, 18, 20, 24, 25, 28, 31
Independent Practice