1. Assignment, red pen, highlighter, textbook, notebook, graphing calculator
U4D2
Have out:
Sketch the following graphs on the same set of axes without using a graphing calculator. Identify the vertex for each parabola.
Bellwork:
f(x) = (x + 1)2 – 2 +2
f(x) = (x + 7)2 x y 4 6 8 10 2 –2 –4 –6 –8
–10
a) f(x) = (x + 7)2 +2
vertex: (–7, 0) +1 +2
f(x) = (x – 8)2 + 5
b) f(x) = (x – 8)2 + 5
vertex: (8, 5) +1
c) f(x) = (x + 1)2 – 2
vertex: (–1, –2) +1
total:

2. Stretch / Compress Parabolas
Take out the
worksheet
Stretch Factor:
Stretch / Compress Parabolas
Graph the parabolas using a graphing calculator. Record any observations you notice about the similarities and differences of each parabola with the parent graph, y = x2. x y 4 6 8 10 2 –2 –4 –6 –8
–10
Parent Graph: y = x2 x -3 -2 -1 1 2 3 y 9 4 1 1 4 9

3. Observations: y x 10
–10
Skinnier, stretched vertically
Wider, compressed vertically
Skinnier, stretched vertically
Wider, compressed vertically
Skinnier, stretched vertically, and flipped

4. Conclusions:
Given that the general equation of parabolas is y = a(x – h)2 + k, describe what “a” does to the equation:
Stretches or compresses the parabola
If |a| < 1, the graph is ____________________.
vertically compressed
vertically stretched
If |a| > 1, the graph is ____________________.
“a” is always _________, however, it is the ____ in front of “a” that determines the graph’s ___________ (i.e, open _____ or _____).
positive
+/-
orientation up
down

5. Horizontal shift: _____________ Vertical Shift: _______________
Examples: To graph equations in the form y = a(x – h)2 + k, do the following: a)
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
1 to the right
Identify the horizontal and vertical shifts.
Plot the vertex first.
3) Plot the rest of the parabola. Keep in mind that the distance the points are from the vertex will change depending on the stretch factor.
7 down
(1, -7) 2 up x y 4 6 8 10 2 –2 –4 –6 –8
–10
(1, –7)

6. b) c) (–2, 8) (5, 0) 2 to the left 5 to the right
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
8 up
0 up
(–2, 8)
(5, 0) 3
down 4
down x y 4 6 8 10 2 –2 –4 –6 –8
–10 x y 4 6 8 10 2 –2 –4 –6 –8
–10
(–2, 8)
(5, 0)

7. d) e) (–4, –3) (3, –6) 4 to the left 3 to the right
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
3 down
6 down
(–4, –3)
(3, –6) up up x y 4 6 8 10 2 –2 –4 –6 –8
–10 x y 4 6 8 10 2 –2 –4 –6 –8
–10
(–4, –3)
(3, –6)

8. Take out the Parent Graph Toolkits and your graph paper notebooks.f)
2 to the right
Horizontal shift: _____________
Vertical Shift: _______________
Vertex: _________
Stretch |a| = ___ Orientation: ___
5 up
(2, 5)
1/3
down x y 4 6 8 10 2 –2 –4 –6 –8
–10
Finish the rest later!
(2, 5)
Take out the Parent Graph Toolkits and your graph paper notebooks.

9. Vertex (or Graphing) Form
PG – 22
Standard Form
Add to your notes:
A quadratic equation in standard form looks like:
y = ax2 + bx + c
y = 3x2 – 6x – 2
Example:
a = 3
b = – 6
c = – 2
Vertex (or Graphing) Form
A quadratic equation in vertex (or graphing) form looks like:
y = a(x – h)2 + k
The vertex of a parabola “locates” its position on a set of axes. The vertex serves as the LOCATOR POINT for a parabola.
(h, k)
The vertex has the general form:

10. parabolas opening down
PARENT GRAPH TOOLKIT y 10
Name:
Parabolas x y x y -3 -2 -1 1 23 9 4 49
(3, 9) 8
Parent Equation:
y = x2 6 4
(2, 4)
Description of Locator: 2
(1, 1) x
vertex
(h, k) –4 –2 2 4
(0, 0) –2
General Equation: –4
y = a(x – h)2 + k –6
Properties: –8
Open up or down
horizontal shift
–10
Domain:
Range:
y =+/- a(x – h)2 + k
parabolas opening up
{x| x  R}
{y| k ≤ y < }
Wider or narrower
{y| –< y ≤ k}
vertical shift
parabolas opening down

11. PG – 22
a) Use your graphing calculator to verify that the 2 equations y = 3(x – 1)2 – 5 and y = 3x2 – 6x – 2 are the same. ^
<
Yep, they are the same.
(1, – 5)

12. PG – 22
b) Show algebraically that these two equations are equivalent by starting with vertex form and showing step by step how to get the standard form.
y = 3(x – 1)2 – 5 x
– 1
y = 3(x2 – 2x + 1) – 5 x x2
– x
y = 3x2 – 6x + 3 – 5
– 1
– x
+ 1
y = 3x2 – 6x – 2
c) Notice that the value for “a” is 3 in both forms of the equation, but that the numbers for “b” and “c” are different from the numbers for “h” and “k”. Why do you think the value of “a” would be the same number in both forms of the equation?
3 is still the coefficient after the binomial is squared. The other numbers are changed when like terms are combined.

13. PG – 23
The students in Ms. Speedi’s class had to graph f(x) = x2 – 8x + 7 without using their graphing calculators. Making xy–tables and plotting point–by–point would take forever. They had to find other way.
a) Jessica suggested, “We know how to get the coordinates of the x–intercepts.” What are the coordinates of the x–intercepts for this equation?
0 = x2 – 8x + 7 7
0 = (x – 1)(x – 7)
Set f(x) = 0
– 1
– 7
x – 1 = 0 or
x – 7 = 0
– 8
x = 1
x = 7
(1, 0)
(7, 0)
b) Kevin said, “And parabolas are symmetric, so the vertex is mid–way between them.” How could you find the x–value of the vertex? Find it.
Take the average of the x–values.
= 4
Recall that x = 4 is the line of symmetry.

14. PG – 23
c) Kenya volunteered, “And if we know the x–value, we can find the y–value that goes with it.” Find the y–value.
x = 4
f(4) = (4)2 – 8(4) + 7
The vertex is (4, -9).
f(4) = 16 –
f(4) = -9

15. PG – 23
d) “We’ve got the vertex. Now we can make a sketch!” they all said together. Make a sketch of this function, label the intercepts and the vertex, and draw in the line of symmetry.
x = 4 2 4 6 8 10 y x –2 –4 –6 –8
–10
(0, 7)
f(x) = x2 – 8x + 7
(1, 0)
(7, 0)
(4, –9)

16. PG – 23
e) “How can we be sure we’re right?” asked Jessica. “Couldn’t we write the vertex form equation then work it out to see if we get the original one in standard form?” Enrico suggested. Use the vertex to write the equation in vertex form. Show your work to verify that it is the same as the original by squaring and simplifying.
vertex (4, –9)
y = a(x – h)2 + k
y = (x – 4)2 – 9
y = (x2 – 8x + 16) – 9
y = (x – 4)2 – 9
y = x2 – 8x + 16 – 9
y = x2 – 8x + 7

17. For each function below, do the following:
PG – 24
For each function below, do the following:
(i) Algebraically solve for the x–intercepts.
(ii) Determine the line of symmetry.
(iii) Determine the vertex.
(iv) Graph the parabola and label the intercepts and vertex.
Write the equation in vertex form. Verify that the equation is the same as the standard form equation.
a) p(x) = x2 – 10x + 16
b) f(x) = x2 + 3x – 10

18. a) p(x) = x2 – 10x + 16 x–intercepts: x2 – 10x + 16 = 0
PG – 24 2 4 6 8 10 y x –2 –4 –6 –8
–10
a) p(x) = x2 – 10x + 16
x–intercepts:
p(x) = x2 – 10x + 16
x2 – 10x + 16 = 0
(x – 2)(x – 8) = 0
(2, 0)
x – 2 = 0 or x – 8 = 0
(8, 0)
x = 2 or x = 8
(2, 0) and (8, 0)
line of symmetry:
(5, –9)
= 5
x = 5
vertex form equation:
x = 5
p(x) = (x – 5)2 – 9
vertex:
p(x) = x2 – 10x + 25 – 9
p(5) = (5)2 – 10(5) + 16
p(x) = x2 – 10x + 16
p(5) = – 9
(5, – 9)

19. b) f(x) = x2 + 3x – 10 x–intercepts: x2 + 3x – 10 = 0
PG – 24 2 4 6 8 10 y x –2 –4 –6 –8
–10
–12
–14
f(x) = x2 + 3x – 10
b) f(x) = x2 + 3x – 10
x–intercepts:
(– 5, 0)
(2, 0)
x2 + 3x – 10 = 0
(x + 5)(x – 2) = 0
x + 5 = 0 or x – 2 = 0
x = – 5 or x = 2
(– 5, 0) and (2, 0)
(0, –10)
line of symmetry:
vertex form equation:
vertex:

20. Finish today's assignment:
PG 22, 23, ,
Stretch Factor & PG-24 WS

21. OLD SLIDES

22. c) g(x) = x2 – 4x – 2 x–intercepts: x2 – 4x – 2 = 0 line of symmetry:
PG – 24 2 4 6 8 10 y x –2 –4 –6 –8
–10
c) g(x) = x2 – 4x – 2
x–intercepts:
g(x) = x2 – 4x – 2
x2 – 4x – 2 = 0
(0, –2)
line of symmetry:
(2, –6)
x = 2
x = 2
vertex:
vertex form equation:
g(2) = (2)2 – 4(2) – 2
g(x) = (x – 2)2 – 6
g(2) = – 6
(2, – 6)
g(x) = x2 – 4x + 4 – 6
g(x) = x2 – 4x – 2

23. d) h(x) = –4x2 + 4x + 8 x–intercepts: –4x2 + 4x + 8 = 0
PG – 24 2 4 6 8 10 y x –2 –4 –6 –8
–10
d) h(x) = –4x2 + 4x + 8
(0, 8)
x–intercepts:
h(x) = –4x2 + 4x + 8
–4x2 + 4x + 8 = 0
–4(x2 – 1x – 2) = 0
(–1, 0)
(2, 0)
–4(x + 1)(x – 2) = 0
x + 1 = 0 or x – 2 = 0
x = –1 or x = 2
(–1, 0) and (2, 0)
line of symmetry:
vertex form equation:
vertex:

24. Vertex (or Graphing) Form
PG – 22
Vertex (or Graphing) Form
Add to your notes:
A quadratic equation in vertex (or graphing) form looks like:
Horizontal change
+ h shift right (x – h)
– h shift left (x + h)
y = a(x – h)2 + k
|a| > 1 narrow parabola (“contraction”)
Vertical change
|a| < 1 wider parabola (“dilation”)
+ k shift up
opens upward
– k shift down
– opens downward
(h, k)
The vertex has the general form: