Download everything you need to know about parabolas

EVERYTHING YOU NEED TO KNOW ABOUT
PARABOLAS
Vertex format is:
2
f  x   a  x  h  k
Standard format is:
f  x   ax 2  bx  c
The axis of symmetry is at
xh
The axis of symmetry is at
b
x
2a
The vertex is at
 h, k 
The vertex is at
 h, f  h  
Graph shifts:
 k shifts graph up
k shifts graph down
The y-intercept is (0, c)
 x  h
 x  h
shifts RIGHT
shifts LEFT
Graph shifts:
Use conversion formulas to get h and k
b
h
and
k  f  h
2a
ax 2 stretches vertically if a  1
compresses vertically if a  1
 jx 
2
compresses horizontally if j  1
stretches horizontally if j  1
If a is positive, then:
 The parabola opens up
 The vertex is the minimum
 The range is y  k or y  f  h ,
depending on which format you’re
working with
If a is negative, then:
 The parabola opens down
 The vertex is the maximum
 The range is y  k or y  f  h
Domain is always: all real numbers,
regardless of the sign of a
If you pick two points on the parabola that are located at the same height, then the midpoint
between their x-coordinates will be the value of h
The “roots of a quadratic”, the “zeroes of a quadratic”, and the “x-intercepts of a quadratic” all
mean the same thing --- find out what values of x make the equation equal zero.
To find the roots, factor the equation into its original pieces (sometimes called “reverse FOIL”). Set
each factor equal to zero. This breaks the original function into several mini-equations, and
the solutions to those equations are the zeroes of the function.
Because the roots all have the same y-value (zero), then midpoint between the roots is h.
To get a chart of values, take advantage of the parent function relationship and graph shifts.
Here is the parent function’s table:
x x2 f  x 
0 02
0
2
1 1
1
2
2 2
4
2
3 3
9
2
4 4
16
So if you have: f  x    x  1  2 ,
The parent function has been shifted right 1 and up 2.
So all you need to do is shift the parent table’s first and last columns to the right 1 and up 2:
2
x right 1 x 2
01
11
21
31
41
f  x  up 2
2
02
2
1 2
2
2
42
3
2
92
4
2
16  2
0
1
f  x
x
x2
1
2
2
2
3
2
2
6
3
2
11
4
2
18
2
3
4
5
0
1
Notice that the center column is no longer needed.
Notice also that while the x-column goes up by ones, the f(x) column’s values change by adding 1,
then adding 3, then adding 5, then adding 7…. So you could follow this strategy:
Graph the vertex
From vertex, count right 1, up 1, put a dot
From that dot, count right 1, up 3, put a another dot
From there, count right 1, up 5, put a dot
From that point, count right 1, up 7
Etc, etc, for as many points as you need
Then use the parabola’s symmetry to match those points over on the lefthand side.
If a is a negative 1, instead of positive 1, then multiply the 3rd column by -1 and change your
counting to right and down instead of right and up.
If your problem has a value for a that is not 1 or -1, then take the values in the f(x) column from the
parent function, and multiply them by your a to produce the f(x) values for your function.
If you have an a that is not 1, and a left/right shift, and an up/down shift, BE SURE TO DO ANY
RIGHT/LEFT SHIFTS FIRST, then any multiplying, then any up/down shifts. (Think of it as an
application of the order of operations --- inside the parentheses first, then the multiplication by a,
then the addition or subtraction of k.)
If your function is in standard form, the y-intercept can be used in the chart as well. Its coordinates
are (0, c). And since parabolas are symmetric, there will be a matching point at (2h, c).
If your equation is easily factored, use the factors to find the zeroes. This will give you two more
points for your chart. Also, finding the midpoint between them will give you h.