This summary combines slides from Melanie Tomlinson and Morrobea on the topic of parabolas. The key points covered include: - The geometric definition of a parabola as the set of all points equidistant from a fixed point (the focus) and fixed line (the directrix). - Parabolas can be represented using various equation forms including vertex form, standard form, and general form. - Methods for graphing parabolas by identifying features like the vertex, axis of symmetry, x-intercepts, focus, and directrix. - Applications of parabolas to model real-world situations like searchlights and radio telescopes.

1. This is the combination of
slides from two people which
are
Melanie Tomlinson and
Morrobea.

3. Geometric Definition of a Parabola: The collection of
all the points P(x,y) in a a plane that are the same
distance from a fixed point, the focus, as they are from
a fixed line called the directrix.
y
Focus
(0,p)
P
Verte
x
(h,k)
x

4. •Parabola is a
Quadratic
function

5. Quadratic Function
•
A function of the form y=ax2+bx+c
where a≠0 making a u-shaped graph
called a parabola.
Example quadratic equation:

6. Vertex• The lowest or highest point
of a parabola.
Vertex
Axis of symmetry• The vertical line through the vertex of the
parabola.
Axis of
Symmetry

7. Vertex Form Equation
•
•
•
•
y=a(x-h)2+k
If a is positive, parabola opens up
If a is negative, parabola opens down.
The vertex is the point (h,k).
The axis of symmetry is the vertical line
x=h.
Don’t forget about 2 points on either side
of the vertex! (5 points total!)

8. Vertex Form
Each function we just looked at can be written in the
form (x – h)2 + k, where (h , k) is the vertex of the
parabola, and x = h is its axis of symmetry.
(x – h) 2 + k – vertex form
Equation
Vertex
Axis of Symmetry
y = x2 or
y = (x – 0)2 + 0
( 0 , 0)
x=0
y = x2 + 2 or
y = (x – 0)2 + 2
( 0 , 2)
x=0
y = (x – 3)2 or
y = (x – 3)2 + 0
( 3 , 0)
x=3

9. Example 1: Graph y = (x + 2)2 + 1
Analyze y = (x + 2)2 + 1.
•
• Step 1 Plot the vertex (-2 , 1)
• Step 2 Draw the axis of symmetry, x = -2.
• Step 3 Find and plot two points on one side
, such as (-1, 2) and (0 , 5).
• Step 4 Use symmetry to complete the graph,
or find two points on the
• left side of the vertex.

10. As you can plainly see
the distance from the
focus to the vertex is a
and is the same distance
from the vertex to the
directrix! Neato!
y
Focus
(0,p)
p
2p
Verte
x
(h,k)
Directrix y = -p
And the
equation
is…
x = 4 py
2
p
x

11. y
y
Directrix = a
Verte
x
(h,k)
Focus
(0,-p)
And the
equation
is…
x = −4 py
2
p
x
p

12. y
2p
x
Directrix = −a
p
p
Verte
x
(h,k)
And the
equation
Focus
(p,0)
y = 4 px
2
x

13. y
p
p
Vertex
Focus (p,0)
And the
equation
x
Directrix = a
(h,k)
x
y = −4 px
2

14. Your Turn!
• Analyze and Graph:
y = (x + 4)2 - 3.
(-4,-3)

15. Example 2: Graph
y= -.5(x+3)2+4
•
•
•
•
a is negative (a = -.5), so parabola opens down.
Vertex is (h,k) or (-3,4)
Axis of symmetry is the vertical line x = -3
Table of values
x y
-1 2
Vertex (-3,4)
-2 3.5
(-4,3.5)
(-2,3.5)
-3 4
-4 3.5
(-5,2)
(-1,2)
-5 2
x=-3

16. Try this one!
y=2(x-1)2+3
• Open up or down?
• Vertex?
• Axis of symmetry?
• Table of values with 4 points (other
than the vertex?

17. (-1, 11)
(3,11)
X=1
(0,5)
(2,5)
(1,3)

18. Intercept Form Equation
y=a(x-p)(x-q)
•
•
•
•
•
The x-intercepts are the points (p,0) and (q,0).
+
The axis of symmetry is the vertical line x= p 2 q
p +q
The x-coordinate of the vertex is 2
To find the y-coordinate of the vertex, plug the xcoord. into the equation and solve for y.
If a is positive, parabola opens up
If a is negative, parabola opens down.

19. Example 3: Graph y=-(x+2)(x-4)
• Since a is negative,
•
•
parabola opens down.
The x-intercepts are (2,0) and (4,0)
To find the x-coord. of
p+q
the vertex, use
2
x=
•The axis of symmetry
is the vertical line x=1
(from the x-coord. of
the vertex)
(1,9)
−2+4 2
= =1
2
2
• To find the y-coord.,
plug 1 in for x.
y = −(1 + 2)(1 − 4) = −(3)(−3) = 9
(-2,0)
(4,0)
x=1

20. Now you try one!
y=2(x-3)(x+1)
• Open up or down?
• X-intercepts?
• Vertex?
• Axis of symmetry?

21. x=1
(-1,0)
(3,0)
(1,-8)

22. Changing from vertex or intercepts
form to standard form
• The key is to FOIL! (first, outside, inside,
last)
• Ex: y=-(x+4)(x-9)
=-(x2-9x+4x-36)
=-(x2-5x-36)
y=-x2+5x+36
Ex: y=3(x-1)2+8
=3(x-1)(x-1)+8
=3(x2-x-x+1)+8
=3(x2-2x+1)+8
=3x2-6x+3+8
y=3x2-6x+11

23. Challenge Problem
• Write the equation of the graph in vertex form.
y = 3( x + 2) 2 − 4

24. Graphing an Equation of a Parabola
Standard Equation of a Parabola (Vertex at Origin)
x = 4 py
2
x = 12 y
2
( 0, p )
focus
y= −p
directrix
y = −3
( 0, 3)

25. Graphing an Equation of a Parabola
Standard Equation of a Parabola (Vertex at Origin)
y = 4 px
2
y = 12 x
2
( p, 0 )
focus
x= −p
directrix
x = −3
( 3, 0)

26. Graphing an Equation of a Parabola
Graph the equation. Identify the focus and directrix of the
parabola.
1. x = 2 y
2
1
4p = 2 p =
2
 1
focus:  0, 
 2
1
directrix: y = −
2

27. Graphing an Equation of a Parabola
Graph the equation. Identify the focus and directrix of the
parabola.
2. y = 16 x
2
4 p = 16 p = 4
focus: ( 4, 0 )
directrix: x = − 4

28. Graphing an Equation of a Parabola
Graph the equation. Identify the focus and directrix of the
parabola.
1
3. x = − y
4
1
1
4p = − p = −
16
4
2
1

focus:  0, − 
16 

1
directrix: y =
16

29. Graphing an Equation of a Parabola
Graph the equation. Identify the focus and directrix of the
parabola.
4. y = −4 x
2
4 p = −4 p = −1
focus: ( − 1, 0 )
directrix: x = 1

30. Writing an Equation of a Parabola
Write the standard form of the equation of the parabola
with the given focus and vertex at (0, 0).
2
5. 0, 1
(
)
x = 4( 1 ) y
2
 1 
6.  − , 0 
 2 
y =
2
( )x
1
4−
2
x = 4 py
x = 4y
2
y = 4 px
2
y = −2 x
2

31. Writing an Equation of a Parabola
Write the standard form of the equation of the parabola
with the given focus and vertex at (0, 0).
7. ( − 2, 0 )
y = 4( − 2) x
2
 1
8.  0, 
 4
x = 4(
2
y = 4 px
2
y = −8x
2
x = 4 py
2
1
4
)y
x =y
2

32. Modeling a Parabolic Reflector
9. A searchlight reflector is designed so
that a cross section through its axis is a
parabola and the light source is at the
focus. Find the focus if the reflector is 3
feet across at the opening and 1 foot deep.
(1x5) = 4 p (1)
.
y
2 2
(1.5, 1)
2.25 = 4 p
2.25 9
225
p=
=
4
400 16

33. Notes Over 10.2
Modeling a Parabolic Reflector
10. One of the largest radio telescopes has
a diameter of 250 feet and a focal length
of 50 feet. If the cross section of the radio
telescope is a parabola, find the depth.
2
250
x = 4 py
2
x = 4( 50 ) y
x = 200 y
2
125 = 200 y
2
2
= 125
15,625 = 200 y
y = 78.1 ft

34. General Form of any Parabola
Ax + By + Cx + Dy + E = 0
2
2
*Where either A or B is zero!
* You will use the “Completing the
Square” method to go from the
General Form to Standard Form,

35. Graphing a Parabola: Use completing the square
to convert a general form equation to standard
conic form
General form
y2 - 10x + 6y - 11 = 0
y2 + 6y + 9 = 10x + 11 + _____
9
(y + 3)2 = 10x + 20
(y + 3)2 = 10(x + 2)
Standard form
aka: Graphing form
(y-k)2 = 4p(x-h)

36. References
• Melanie Tomlinson, M. 2012.
•
http://www.slideshare.net/melanieitomlinson/parabola-14893256?qid=170dff68-7
. Accessed on 06 March 2014
Morrobea. 2013.
http://www.slideshare.net/morrobea/52-solve-quadratic-equations-by-graphingver
http://www.slideshare.net/morrobea/52-solve-quadratic-equations-by-graphingve
. Accessed on 06 March 2014