1. The powers of General Form

2. Probe
Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.
a) Find and correct the odd-one-out.
b) Name the 5 forms x y -1 5 1 8 2 9 3 4

3. Probe
Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.
a) Find and correct the odd-one-out.
b) Name the 5 forms x y -1 5 1 8 2 9 3 4
mapping rule
equation in general form
equation in transformational form
graphical (parabola)
table of values

4. General Form to Transformational Form
Let’s look at an example. We will take the quadratic function given in general form: y = 2x2 + 12x – 4 and turn it into transformational form: 1
We know that in transformational form the coefficient of ‘x’ is 1
STEP 1: Divide every term by a.

5. General Form to Transformational Form
Let’s look at an example. We will take the quadratic function given in general form: y = 2x2 + 12x – 4 and turn it into transformational form:
STEP 2: Move the non-x term over

6. x2 Completing the Square x x x x x x
We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2
So far we have: x2 + 6x
This looks like: x2 x x x x x x

7. x2 x2 Oops… Completing the Square x x
The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape... x2 x
Oops…

8. x2 x2 Completing the Square x x x x x x x Try again… Closer….
We just need to complete this square… x x x
We were missing nine 1×1 squares.

9. x2 Completing the Square x
This looks promising: We have a square with length and width dimensions of (x + 3).
We just had to add 9, or 32. x
But how can we justify this “just add 9”?
Add it to both sides of the equation!
But why 32?
It’s half of the coefficient of x, squared.

10. General Form to Transformational Form
Back to our example. We will take the quadratic function given in general form: y = 2x2 + 12x – 4 and turn it into transformational form:
STEP 3: Complete the square (take half the coefficient of x, square it and add it to both sides)

11. General Form to Transformational Form
Back to our example. We will take the quadratic function given in general form: y = 2x2 + 12x – 4 and turn it into transformational form:
STEP 4: Factor out the 1/a term on the left hand side

12. General Form to Transformational Form
Back to our example. We will take the quadratic function given in general form: y = 2x2 + 12x – 4 and turn it into transformational form:
Therefore, the transformational form of the quadratic function y = 2x2 + 12x – 4 is:
Now we know that its VS = 2, HT = –3, and VT = –22
From this we know its vertex, range, axis of symmetry, etc
vertex is (–3, –22)
range is {y≥ –22, y  R}
axis of symmetry is x = –3

13. Practice
1. Complete the square to find the vertex of the function y = 2x2 – 8x + 2. Sketch its graph. 2. Complete the square to find the range of the function y = –x2 – 5x +1? 3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

14. Practice: Solutions
Complete the square to find the vertex of the function y = 2x2 – 8x Sketch its graph.
Vertex: (2, –6)

15. Practice: Solutions
2. Complete the square to find the range of the function y = –x2 – 5x +1?
range is {y ≤ 7.25, y  R}

16. Practice: Solutions
3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

17. A shortcut…eventually
But completing the square is a lot of work…. Let’s find a shortcut to finding vertex ect. by completing the square just ONE more time, but with the general equation
STEP 1: Divide every term by a.
STEP 2: Move the non-x term over
STEP 3: Complete the square (take half the coefficient of x, square it and add it to both sides)
STEP 4: Factor out the 1/a term on the left hand side

18. A shortcut…eventually
STEP 1: Divide every term by a.
STEP 2: Move the non-x term over
STEP 3: Complete the square (take half the coefficient of x, square it and add it to both sides)
STEP 4: Factor out the 1/a term on the left hand side

19. Here’s the shortcut
We now see that ANY general quadratic equation y = ax2 + bx + c can be written as
This may look complicated, but is VERY helpful…
from this we can see that the HT for any general quadratic is
For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6?
To complete the square on this takes time. But…

20. Here’s the shortcut
We now see that ANY general quadratic equation y = ax2 + bx + c can be written as
This may look complicated, but is VERY helpful…
from this we can see that the VT for any general quadratic is
For example: What is the vertex of the function y = –2x2 + 4x + 6?
To complete the square on this takes time. But…
vertex (1, 8)

21. Here’s the shortcut
We now see that ANY general quadratic equation y = ax2 + bx + c can be written as
This may look complicated, but is VERY helpful…
from this we can see that the VS for any general quadratic is a.
Since the VS = –2 we can graph from the vertex (1, 8):
Over 1 down 2
Over 2 down 8
Over 3 down 18
For example: Graph the function y = –2x2 + 4x + 6.
To complete the square on this takes time. But…

22. Shortcut within a shortcut
Instead of memorizing the formula for the y-coordinate of the vertex (VT):
we can calculate it using the general form of the equation and the x-coordinate of the vertex (HT):
For example: What is the maximum value of the function y = –2x2 + 4x + 6?
Max value is y = 8.

23. General form shortcuts: Practice
For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.

24. General form shortcuts: Practice Solutions
vertex (1, –2), VS = 1/4
axis of symmetry: x = 1 range: {y ≥ 2, y  R}

25. General form shortcuts: Practice Solutions
vertex (–4, –6), VS = 3
axis of symmetry: x = –4 range: {y ≥ –6, y  R}

26. Solving for x given y using general form
Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2.
Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation.
A quadratic equation in general form looks like this: ax2 + bx + c = 0 where a ≠ 0.
Solving a quadratic equation for x is also called finding its roots or finding its zeros.

27. Solving for x given y using general form
NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example.
Example 1: Give the quadratic equation obtained from the function y = x2 – 2x – 15 when y = –12.
Although the quadratic function had coefficients a = 1, b = –2, and c = –15, subbing in −12 for y gives the quadratic equation with coefficients a = 1, b = –2, and c = –3
Solution:
−12 = x2 – 2x – 15
0 = x2 – 2x – 3

28. Solving for x given y using general form
NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example.
Example 2: Give the quadratic equation obtained from the function y = x2 – 2x – 15 when y = 0.
Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15.
Solution:
0 = x2 – 2x – 15
While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.

29. Solving for x-intercepts using general form
Example 3: Find the x-intercepts of the quadratic function y = x2 – 2x – 15
Solution:
0 = x2 – 2x – 15
…but now what?
Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form…
So lets try that!

30. Solving for x-intercepts using general form
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the squarroot of both sides
STEP 5: Isolate x in both equations
Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.

31. Solving for x-intercepts using general form Practice
Find the roots of the following quadratics:
Not worth it!
Wait for the
shortcut!

32. A shortcut…eventually
But this is a lot of work…. Let’s find a shortcut to finding x-intercepts by doing this just ONE more time, but with the general equation.
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations 32

33. A shortcut…eventually
STEP 1: Write in transformational form by completing the square
STEP 2: Set f(x) = y = 0
STEP 3: Simplify left-hand side 33

34. A shortcut…eventually
STEP 4: Take the square-root of both sides
STEP 5: Isolate x in both equations
This is the shortcut, the Quadratic Root Formula! 34

35. Solving for x-intercepts using general form The short-cut
Back to our example: Find the x-intercepts of the graph of the function f(x) = x2 – 2x – 15
Solution:
Instead of those 5 steps, let’s apply the Quadratic Root Formula.

36. Solving for x given y using general form
The quadratic root formula can be used to find x-values other than the x-intercepts
For the function f(x) = x2 – 2x – 15, find the values of x when y = −12:
Solution:
–12 = x2 – 2x – 15
0 = x2 – 2x – 3.
Now use the quadratic root formua with a = 1, b = –2 and c = –3

37. Solving for x-intercepts using general form Practice
Find the roots of the following quadratics:
Use the
quadratic
root formula!

38. Finding roots from General Form: Practice Solutions

39. Finding roots from General Form: Practice Solutions

40. Finding roots from General Form: Practice Solutions
Find the x-intercepts of the following quadratic functions:
Answers:

41. Yet another method of finding roots: Factoring
As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster!
The zero property:
If (s)(t) = 0 then s = 0, or t = 0, or both. (This only works when the product is 0.)
So if we can get the quadratic function in the form: y = (x – r1)(x – r2), (called factored form) then when y = 0 we know that:
(x – r1) = 0, or (x – r2) = 0, or both.
In other words, x = r1 or x = r2 or both.

42. Factoring to find the roots
Ex. Find the x-intercepts of the graph of the function f(x) = x2 – 2x – 15
Answer when the numbers are multiplied
Answer when the numbers are added
We need to factor x2 – 2x – 15. We need 2 numbers whose product is –15 and whose sum is –2
(There is never more than one pair of numbers that work!)
___ × ___ = –15
___ + ___ = – 2 -5 3 -5 3
So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3
So the factored form of f(x) = x2 – 2x – 15 is f(x) = (x – 5)(x + 3)
So the x-intercepts of the graph are (5, 0) and (–3, 0)

43. Factoring
When factoring a quadratic equation where a = 1, find two numbers that multiply to give c and add to give b.
By the way:
(0) = x2 + 10x – 24
24 = x2 + 10x
= x2 + 10x + 25
49 = (x + 5)2
±7 = x + 5
x = – or x = –5 – 7
x = or x = –12
Ex. Find the roots of y = x2 + 10x – 24 by factoring.
___ x ___ = -24
___ + ____ = 10 12 -2 12 -2 or

44. Practice 1. Solve the following equations by factoring.
a) 0 = x2 + 2x +1
c) 0 = x2 + 2x – 24
b) 0 = x2 + 5x +4
d) 0 = x2 – 25
2. Find the x- and y- intercepts of the following quadratics.
a) f(x) = 2x2 +3x + 1
c) f(x) = x2 – 5x -14
b) f(x) = 3x2 – 7x + 2
d) y = 2(x – 4)2 – 32
Answers
1a. x = –1 b. x = –4 or –1 c. x = –6 or 4 d. x = –5 and 5
2a. (–1, 0) (–0.5, 0) (0, 1) b. (1/3, 0) (2, 0) (0, 2)
c. (–2, 0) (7, 0) (0, –14) d. (0, 0) (8, 0)