1. “Exploring Quadratic Functions”

2. Quadratic functions around you

3. Quadratic Functions (what you need to know…
Quadratic Functions (what you need to know….you’ll come across them again and again):
Functions with the form y=ax2+bx+c are called quadratic functions and their graphs have a parabolic shape
When we solve ax2+bx+c=0 we look for values of x that are x-intercepts (because we have y=0)
The x-intercepts are called the solutions or roots of a quadratic equation
A quadratic equation can have two real solutions, one real solution, or no real solutions

4. How does the function look like?
For example: y= -x2 - 2x + 8.
The graph has:
Vertex: (-1, 9)
Roots: (-4, 0) (2, 0)

5. Maximum and Minimum Points
a is positive, therefore the parabola opens upward, and the vertex is the minimum point.
a is negative, therefore the parabola opens downward, and the vertex is the maximum point.
(-1, 2)
Vertex
Turning Point
Vertex
Turning Point
(1, -2)

6. Axis of Symmetry
The Axis of Symmetry of a parabola is the line that splits the parabola in half lengthwise. The Axis of Symmetry always goes through the Vertex of the parabola.
Let’s look at some graphs.
Axis of
Symmetry
x = -1
Axis of
Symmetry
x = 1

7. Quadratic Functions Forms
Function: (standard form)
Vertex Form:

8. Let’s review what you have learnt
Let’s review what you have learnt. Show that g represents a quadratic function. Identify a, b, and c.

9. Show that g represents a quadratic function. Identify a, b, and c.

10. Show that g represents a quadratic function. Identify a, b, and c.

11. Let’s take a look at the graph. “Parts of a Parabola”
Axis of Symmetry
(Line of Symmetry) LOS:
The line that divides the parabola into two parts that are mirror images of each other.
Vertex:
Either the lowest (minimum) or highest (maximum) point.

12. Let’s look at some graphs.
What is the vertex, max or min, and los?

13. Vertex Form:

14. Let’s look at another graph (transformation)
What is the vertex, max or min, and los?

15. Vertex Form:

16. Here’s another one…
What is the vertex, max or min, and los?

17. Vertex Form:

18. Quadratic Function Transformations Quick Summary
Vertex Form:
h is horizontal translation
(‘-’ to the right and ‘+’ to the left)
k is vertical translation
(‘-’ move down and ‘+’ move up)

19. Quadratic Functions (Re-cap…here we go again):
Functions with the form y=ax2+bx+c are called quadratic functions and their graphs have a parabolic shape
When we solve ax2+bx+c=0 we look for values of x that are x-intercepts (because we have y=0)
The x-intercepts are called the solutions or roots of a quadratic equation
A quadratic equation can have two real solutions, one real solution, or no real solutions

20. Solving Quadratic Functions

21. Properties of Solving Quadratic Equations

22. …by Quadratic Formula
You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions (…not for now).
The quadratic formula gives the solutions of
ax2 + bx + c = 0 when it is not easy to factor the quadratic (or complete the square).

23. Example:
Find the zeros of f(x)= 2x2 – 16x + 27 using the Quadratic Formula.
2x2 – 16x + 27 = 0
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 2 for a, –16 for b, and 27 for c.
Simplify.
Write in simplest form.

24. Solve x2 – 9x + 20 = 0. Show your work.
Example:
Solve x2 – 9x + 20 = 0. Show your work.
Method 4 Solve using the Quadratic Formula.
1x2 – 9x + 20 = 0
Identify a, b, c.
Substitute 1 for a, –9 for b, and 20 for c.
Simplify.
Write as two equations.
x = 5 or x = 4
Solve each equation.

25. Method 4 Solve using the Quadratic Formula.
Now you try it…
x2 + 7x + 10 = 0
Method 4 Solve using the Quadratic Formula.
1x2 + 7x + 10 = 0
Identify a, b, c.
Substitute 1 for a, 7 for b, and 10 for c.
Simplify.
Write as two equations.
x = –5 or x = –2
Solve each equation.

26. …by Quadratic Formula
The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

27. Discriminant
Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!

28. Example: Analyzing Quadratic Equations by using the Discriminant
Find the type and number of solutions for the equation.
x = 12x
x2 – 12x + 36 = 0
b2 – 4ac
(–12)2 – 4(1)(36)
144 – 144 = 0
b2 – 4ac = 0
The equation has one real solution.

29. Example: Analyzing Quadratic Equations by using the Discriminant
Find the type and number of solutions for the equation.
x = 12x
x2 – 12x + 40 = 0
b2 – 4ac
(–12)2 – 4(1)(40)
144 – 160 = –16
b2 –4ac < 0
The equation has two distinct non-real solutions.

30. Example: Analyzing Quadratic Equations by using the Discriminant
Find the type and number of solutions for the equation.
X = 13x
x2 – 13x - 30 = 0
b2 – 4ac
(–13)2 – 4(1)(-30)
= 289
b2 – 4ac > 0
The equation has two distinct real solutions.

31. Solving Quadratic Equations by Graphing (some common terms)
In a quadratic equation y=ax2+bx+c, ax2 is the quadratic term, bx is the linear term, and c is the constant term
The axis of symmetry is a line that divides a parabola into two equal parts that would match exactly if folded over on each other
The vertex is where the axis of symmetry meets the parabola
The roots or zeros (or solutions) are found by solving the quadratic equation for y=0 or looking at the graph

32. …by Graphing Graph with definitions shown:
Three outcomes for number of roots:
Two roots
One root:
No roots:

33. …by Graphing
For y= -x2 -2x + 8 graph the equation, find the vertex, and find the solutions of the equation.
Vertex:
x= -b/2a
x= -(-2)/2(-1)
x= 2/(-2)
x= -1
Solve for y:
y= -x2 -2x + 8
y= -(-1)2 -2(-1) + 8
y= -(1)
y= 9
Vertex is (-1, 9)

34. …by Graphing Find the roots (solutions) for the Problem:
-x2 -2x + 8 = 0
(-x + 2)(x + 4) = 0
-x + 2 = x + 4 = 0
-x = x = -4
x = 2
(2, 0) and (-4, 0) are the roots.

35. …by Graphing Now you try it… Solve and show your work. –14 + x2 = 5x
y = x2 – 5x – 14
Vertex:
x= -b/2a
x= -(-5)/2(1)
x= 5/2
Solve for y:
y = x2 -5x – 14
y = (5/2)2 –(5)(5/2) – 14
y = 25/4 - 25/2 – 14
y = -81/4
Vertex is (5/2, -81/4)

36. …by Graphing Find the roots (solutions) for the Problem:
x2 -5x - 14 = 0
(x + 2)(x - 7) = 0
x + 2 = x - 7 = 0
x = x = 7
(-2, 0) and (7, 0) are the roots.

37. …by Completing the Square
The way to complete a square for x2 + bx + ? is to take ½ x b and then square it
So for x2 + 6x + ? :
½ (6) = = Therefore, the blank should be 9.
If the coefficient of x2 is not 1, you must divide the equation by that coefficient before completing the square
Some roots will be irrational or imaginary numbers

38. The Aim of Completing the Square
… is to write a quadratic function as a perfect square. Here are some examples of perfect squares!
x2 + 6x + 9
x2 - 10x + 25
x2 + 12x + 36
Try to factor these (they’re easy).

39. Perfect Square Trinomials
=(x+3)2
=(x-5)2
=(x+6)2
x2 + 6x + 9
x2 - 10x + 25
x2 + 12x + 36
Can you see a numerical connection between …
6 and 9 using 3
-10 and 25 using -5
12 and 36 using 6

40. The Perfect Square Connection
For a perfect square, the following relationships will always be true … x2 + 6x + 9 x2 - 10x + 25
Half of these values
squared
… are these values

41. The Perfect Square Connection
In the following perfect square trinomial, the constant term is missing. Can you predict what it might be?
X2 + 14x + ____
Find the constant term by squaring half the coefficient of the linear term.
(14/2)2
X2 + 14x + 49

42. Perfect Square Trinomials
Create perfect square trinomials.
x2 + 20x + ___
x2 - 4x + ___
x2 + 5x + ___
100 4
25/4

43. …by Completing the Square
Solve the following
equation by
completing the
square:
Step 1: Move the constant term (i.e. the number) just a li….ittle bit further to right
x2 + 8x = 0

44. …by Completing the Square
Step 2: Find the term that completes the square. Add that term to the expression.
Step 3: Put the opposite sign of the term found in step 2.
Step 4: Factor and simplify the expression.
x2 + 8x + (½(8)) = 0
x2 + 8x = 0
x2 + 8x = 0
(x + 4) = 0

45. …by Completing the Square
Step 5:
Solve for x.
(x + 4) = 0
(x + 4)2 = 36
x + 4 =
x + 4 = 6
x = or x =
x = or x = 2
Solving is a simple process …

46. …by Completing the Square
2x2 - 7x = 0
2(x2 – ( 7/2)x) = 0
2(x2 – ( 7/2)x + (-7/4)2) – 2 (-7/4) = 0
2(x – ( 7/4) )2 – 2 (-7/4) = 0
2(x – ( 7/4) )2 – 49/ = 0
2(x – ( 7/4) )2 + 47/8 = 0

47. …by Completing the Square
2(x – ( 7/4) )2 + 47/8 = 0
2(x – ( 7/4) )2 = - 47/8
(x – ( 7/4) )2 = - 47/16
In this case we cannot solve for x due to the square root of a negative number, thus no real solution (see Desmos & Discriminant).
Solve for x: