Algebra 2 Chapter 9 Guideline to Graphing Quadratic Functions
In Chapter 9, the main focus is how to graph quadratic equations. These types of
quadratic equations are known as parabolas. The following key terms will be essential for
this chapter:
minimum value- the lowest point of the graph (more in-depth in this study guide)
maximum value- the highest point of the graph (more in-depth in this study guide)
standard form of a parabolatransformation- the alteration of a graph
translation- the movement of a graph going up/down or left/right without rotation
vertex of a parabola- the point
where the graph crosses the line of symmetry
x-intercepts- the x-coordinates of the points where a graph crosses the x-axis
Graphing Parabolas
For this part of the study guide, we will learn how to graph
,
and
, which are all different translations of parabolas.
Consider the equation
and its graph
Now look at
and
y
–5
,
y
10
10
5
5
5
x
–5
5
x
The transformation of those two parabolas was caused by the shrinking or stretching of it,
depending on the values. The following is a theorem on how they stretch or shrink:
 If
, the graph is stretched vertically
 If
, the graph is shrunk vertically
 If in any circumstance
, the graph is reflected across the x-axis.
If you were trying to find the vertex for any of those types of graphs, it would be always
located at the origin (0, 0). The line of symmetry for these types would be x = 0.
However, another equation for the parabola will provide a different vertex and line of
symmetry.
y
y
10
10
5
5
–5
5
x
x
–5
(0, –2)
(–3, –2)
Equation:
Vertex:
Line of symmetry:
Equation:
Vertex:
Line of symmetry:
What do you notice about the two graphs? It has shifted to the left. Now since the vertex
was
, the original equation would’ve been
.
Think about it, if the vertex was
instead, what would the equation of the parabola
be? It would be
, in which the parabola shifted to the right by 3 from its
origin.
We have now graphed
and
. How would the graph of
look like? We are going to investigate what happens when you give
a value other than 0.
Consider the two graphs:
vertex and line of symmetry.
and
. We are going to determine the
y
y
10
10
Equation:
Vertex:
Line of symmetry:
5
–5
Equation:
Vertex:
Line of symmetry:
5
5
x
5
x
If you look at the graphs on the previous page, you will notice that the graph from the
origin, shown on the left, had two different translations. It goes to the right by 3 and up
by 2.
How to Determine the Minimum or Maximum Value of a Parabola
We have now learned how to graph parabolas in three different types of equations. The
next big question when analyzing a parabola is to determine the minimum or maximum
value of it. To determine it we identify the a of the equation and take into consideration
the following:
 If
, then
is the lowest point of the graph, and is the minimum value of
the function.
 If
, then
is the highest point of the graph, and is the maximum value
of the function.
Let’s take the equation
for example. Does this graph have a minimum or
maximum value? Since the coefficient in front of the parentheses is 1, then
, which
means that this equation has a minimum value of 2. The graph for this equation is on the
previous page.
Quadratic Functions in Standard Form Transformed into Vertex Form
Remember the quadratic equation
and the vertex form equation
? There is actually a transition from getting from standard form to
vertex form.
Let’s put
terms from the third one.
into vertex form. First thing to do is to separate the 1st two
Separation of terms
Factor out a -2 from the parentheses.
Complete the square.
and
Final equation in vertex form.
If we were to determine the vertex, line of symmetry, and maximum/minimum value, it
would be the following:
Vertex:
Line of symmetry:
Maximum value:
Since the equation has a -2 for it’s coefficient in front of the parentheses, the graph will
be an upside down parabola with a maximum value. Remember that identifying those 3
are very essential to analyzing a parabola.
The X-Intercepts of a Parabola
In Chapter 8, we learned how to determine the nature of solutions of a quadratic equation
by using the discriminant
. Let’s review what the nature of solutions are:

Two real solutions

One real solution

No real solutions/Complex solutions
Now based on that, we will be able to determine the number of x-intercepts in a parabola.
If an equation yields two real solutions, then you will have two x-intercepts. If there’s
only one solution, then there is only one x-intercept. No real solutions would result in
any x-intercepts.
Example: Find the x-intercepts in the equation
.
For this problem you use the discriminant to determine how many x-intercepts you will
get OR you can just go straight ahead and use techniques for solving for x. Just for the
study guide, I will first use the discriminant and then solve for x.
;
,
, and
Using the discriminant to plug-in
Since
, then there will be two real solutions
Now I’m going to prove that there are in fact two real solutions
This is factorable using three-term factoring.
Set to 0 and factor
Principle of zero products
and
There is two solutions after all.
becomes 0