In order to sketch graphs of the form, f(x)=ax2+qf(x)=ax2+q, we need to determine five characteristics:
 sign of aa
 domain and range
 turning point
 yyintercept
 xxintercept
For example, sketch the graph of g(x)=12x23g(x)=12x23. Mark the intercepts, turning point and axis of symmetry.
Firstly, we determine that a<0a<0. This means that the graph will have a maximal turning point.
The domain of the graph is {x:x∈R}{x:x∈R} because f(x)f(x) is defined for all x∈Rx∈R. The range of the graph is determined as follows:
x
2
≥
0

1
2
x
2
≤
0

1
2
x
2

3
≤

3
∴
f
(
x
)
≤

3
x
2
≥
0

1
2
x
2
≤
0

1
2
x
2

3
≤

3
∴
f
(
x
)
≤

3
(21)Therefore the range of the graph is {f(x):f(x)∈(∞,3]}{f(x):f(x)∈(∞,3]}.
Using the fact that the maximum value that f(x)f(x) achieves is 3, then the yycoordinate of the turning point is 3. The xxcoordinate is determined as follows:

1
2
x
2

3
=

3

1
2
x
2

3
+
3
=
0

1
2
x
2
=
0
Divide both sides by

1
2
:
x
2
=
0
Take square root of both sides:
x
=
0
∴
x
=
0

1
2
x
2

3
=

3

1
2
x
2

3
+
3
=
0

1
2
x
2
=
0
Divide both sides by

1
2
:
x
2
=
0
Take square root of both sides:
x
=
0
∴
x
=
0
(22)The coordinates of the turning point are: (0;3)(0;3).
The yyintercept is obtained by setting x=0x=0. This gives:
y
i
n
t
=

1
2
(
0
)
2

3
=

1
2
(
0
)

3
=

3
y
i
n
t
=

1
2
(
0
)
2

3
=

1
2
(
0
)

3
=

3
(23)The xxintercept is obtained by setting y=0y=0. This gives:
0
=

1
2
x
i
n
t
2

3
3
=

1
2
x
i
n
t
2

3
·
2
=
x
i
n
t
2

6
=
x
i
n
t
2
0
=

1
2
x
i
n
t
2

3
3
=

1
2
x
i
n
t
2

3
·
2
=
x
i
n
t
2

6
=
x
i
n
t
2
(24)which is not real. Therefore, there are no xxintercepts which means that the function does not cross or even touch the xxaxis at any point.
We also know that the axis of symmetry is the yyaxis.
Finally, we draw the graph. Note that in the diagram only the yintercept is marked. The graph has a maximal turning point (i.e. makes a frown) as determined from the sign of a, there are no xintercepts and the turning point is that same as the yintercept. The domain is all real numbers and the range is {f(x):f(x)∈(∞,3]}{f(x):f(x)∈(∞,3]}.
Draw the graph of y=3x2+5y=3x2+5.
 Step 1. Determine the sign of aa: The sign of aa is positive. The parabola will therefore have a minimal turning point.
 Step 2. Find the domain and range: The domain is: {x:x∈R}{x:x∈R} and the range is: {f(x):f(x)∈[5,∞)}{f(x):f(x)∈[5,∞)}.
 Step 3. Find the turning point: The turning point occurs at (0,q)(0,q). For this function q=5q=5, so the turning point is at (0,5)(0,5)
 Step 4. Calculate the yintercept: The yintercept occurs when x=0x=0. Calculating the yintercept gives:
y
=
3x2+5
yint
=
3(0)2+5
yint
=
5
y
=
3x2+5
yint
=
3(0)2+5
yint
=
5
(25)  Step 5. Calculate the xintercept: The xintercepts occur when y=0y=0. Calculating the xintercept gives:
y
=
3x2+5
0
=
3x2+5
x2
=
35
y
=
3x2+5
0
=
3x2+5
x2
=
35
(26)
which is not real, so there are no xintercepts.  Step 6. Plot the graph: Putting all this together gives us the following graph:
The following video shows one method of graphing parabolas. Note that in this video the term vertex is used in place of turning point. The vertex and the turning point are the same thing.
 Show that if a<0a<0 that the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)∈(∞;q]}{f(x):f(x)∈(∞;q]}.
Click here for the solution
 Draw the graph of the function y=x2+4y=x2+4 showing all intercepts with the axes.
Click here for the solution
 Two parabolas are drawn: g:y=ax2+pg:y=ax2+p and h:y=bx2+qh:y=bx2+q.
 Find the values of aa and pp.
 Find the values of bb and qq.
 Find the values of xx for which ax2+p≥bx2+qax2+p≥bx2+q.
 For what values of xx is gg increasing ?
Click here for the solution