1. GENERAL to Transformational
CHANGING FORMS
GENERAL to Transformational

2. To General Form – Complete the squar
Remember: the general form of a quadratic function is y = ax2 + bx +c
The transformational form is
The standard form is y = a(x-h)2 + k
To GET general form: FOIL and solve for y.
But how do we go FROM general form?

3. To General Form – Complete the squar
Let’s look at an example:
What is the vertex of the quadratic function y = 2x2 + 12x -4?
We know that in transformational form the coefficient of ‘x’ is 1
So STEP 1 is to divide every term by ‘a’
STEP 2 is to move the non-x term over.

4. x2 Completing the Squar x x x x x x
The right-hand side of the equation needs to be a perfect square: (x –h)2
So far we have: x2 +6x
This looks like: x2 x x x x x x

5. x2 x2 Oops… Completing the Squar x x x x x x x
To make this into a perfect square we could… x2 x
Oops…

6. x2 x2 Completing the Squar x x x x x x x x x x x x
To make this into a perfect square we could…
Almost…we need to complete the square! x2 x x x x x x
We were missing nine 1x1 squares.

7. x2 Completing the Squar x So the PERFECT SQUARE is …
BUT…how did we get this? X x2 x X
Added it to the right hand side. This means we must add it to both sides!
We took HALF the coefficient of x and squared it.
STEP 3: take half the coefficient of x, square it and add it to both sides
(x+3)2

8. x2 Ta Da! Completing the Square x So the PERFECT SQUARE is …
BUT…how did we get this? X x2 x
Step 4: Factor out the 1/a term on the left hand side X
We need the left hand side to have brackets.
Ta Da!
(x+3)2

9. Completing the square We started with the quadratic in general form
y = 2x2 + 12x -4. The same function is
in transformational form.
So what is the range of y = 2x2 + 12x -4 ?
Since the equivalent equation shows us a vertex of (-3,-22) with no reflection, the range is

10. Practice going FROM general form
1. What is the vertex of the function y = 2x2 - 8x +2? 2. What is the range of the function y = -x2 -5x +1? 3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

11. Practice going FROM general form
1. What is the vertex of the function y = 2x2 - 8x +2?
Vertex: (2,-6)

12. Practice going FROM general form
2. What is the range of the function y = -x2 -5x +1?
Y-value of the vertex
Since there’s a reflection

13. Practice going FROM general form
3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

14. A shortcut…eventually
If only there was a way to avoid having to complete the square every time to get the vertex!! Maybe there is…
Let’s put the following equations into transformational form.
y = 3x2 + 12x - 6
y = ax2 + bx + c
STEP 1 is to divide every term by ‘a’
STEP 2 is to move the non-x term over.

15. A shortcut…eventually
STEP 3: 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
Who cares?

16. X = -b 2a Here’s the shortcut
We now see that ANY general equation can be written as
We haven’t finished this yet but we already see that the x-value of the vertex is
X = -b 2a
This is VERY important.
For example: What is the axis of symmetry of the function y = 2x2 – 16x +3?
To complete the square on this takes time. But

17. Finding y
Example 2: What is the range of the graph of the function y = -x2 – 4x+ 7?
We know that the x coordinate of the vertex is
But how do I find y?
You glade the x- value!
You
“plug in it, plug it in”
The range is
NOTE: The vertex of ANY quadratic occurs at x = -b/2a and the max or min value is: f(-b/2a)

18. Changing forms (quickly)
Ex. Graph the parabola defined by the function:
f(x) = 3x2 + 12x - 9
If f(x) = 3x2 + 12x – 9, then the vertex is
We know that transformational form is best for graphing. Is it still necessary?
f(-2) = 3(-2)2 +12(-2)-9
f(-2) = 3*4-24-9
f(-2) = -21
NO!
Vertex (-2,-21)

19. Changing forms (quickly)
So f(x) = 3x2 + 12x -9 can now be written:
If only we knew the ‘a’ value…
Since the VS = 3 we can graph from the vertex:
Over 1 up 3
Over 2 up 12
Over 3 up 27

20. Changing Forms (quickly)
What we know:
f(x) = 3x2 + 12x – 9
Axis of symmetry
Range
(and Domain)
y-intercept
Max/min value
But we don’t know the x-intercepts!

21. Finding our Roots
When dealing with quadratics, we will be asked to solve quadratic equations, find the roots or find the zeros of quadratic equations or find the x-intercepts of the parabola. These all mean: solve for x.
Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15
But we can change this into transformational form.
Since there are 2 x terms (an x2 and an x) we cannot undo what’s being done. So, in this form (for now) we’re stuck!
f(x) = x2 - 2x -15
y = x2 - 2x -15
y+15 = x2 - 2x
y = x2 - 2x +1
y +16 = (x-1)2

22. Finding our Roots f(x) = x2 - 2x -15 y +16 = (x-1)2
We know that for any x-intercepts: f(x) =0
y +16 = (x-1)2
0+16 = (x-1)2
To undo what’s being done to ‘x’ we need to take the square root of both sides.
16 = (x-1)2
So the 2 x-intercepts are (5,0) and (-3,0)
Remember: (-4)(-4) =16
Now, this is actually 2 equations in one

23. Who cares? We Do Finding our Roots
We’ve found the roots of a quadratic function but it involved us going back to transformational form. Is there a shortcut?
When we completed the square on the general form we ended up with this
Who cares?
We Do
If we set y= 0 we can come up with a formula to find the roots of a quadratic function. We’ll call it the “Quadratic Root Formula”

24. Finding our Roots
Now, on the left-hand side we need to add fractions, so we need a common denominator.

25. TA DA!!!!! Finding our Roots So if we go back to our original example:
Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15
First set one side equal to 0 (since y = 0 for x-intercepts)
0 = x2 - 2x -15
Now plug in a, b and c into the quadratic root formula
TA DA!!!!!
So the 2 x-intercepts are (5,0) and (-3,0)

26. Yet another method of finding roots
We have solved quadratic equations by
putting it in transformational form and undoing what’s being done
Making one side 0 and using the quadratic root formula
Sometimes we can also factor. This uses the zero property.
If (r)(s) = 0 then EITHER ‘r’ must equal 0 or ‘s’ must equal 0. This only works when the product is 0.
So if we can get the quadratic function in the form (x-r)(x-s)=0 then we know that either x-r = 0 or x –s has to be zero.
This is called FACTORING

27. 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 added
Again, f(x) = 0
0 = x2 -2x -15
We need two numbers whose product is -15 and whose sum is -2 -5 3
___ x ___ = -15
___ + ___ = -2
Answer when the numbers are multiplied -5 3 So
x2 -2x-15=0
(x-5)(x+3)=0
So the 2 x-intercepts are (5,0) and (-3,0)
Either (x-5) = 0
Or (x+3) = 0
x = 5
x = -3

28. 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:
24 = x2 +10x
= x2 +10x +25
49 = (x+5)2
+/- 7 = x + 5
x = or x = -5 -7
x = 2 or x = -12
Ex. Solve for x by factoring
___ x ___ = -24
___ + ____ = 10 12 -2 12 -2 or
x =( )/2 =2
x = (-10-14)/2 = -12

29. 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
d) y= 2(x-4)2 - 32
b) f(x) = 3x2 -7x + 2
Answers
1a. x = -1 b. x = -4 or -1 c. x = -6 or 4 d. x = -5 or +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)