1) The document discusses graphs of quadratic equations in the form of y = ax^2, y = ax^2 + bx, and y = ax^2 + bx + c. 2) It explains that the graph of y = ax^2 is a parabola that is symmetrical over the y-axis and has a minimum point at the vertex (0,0). The factor a stretches or compresses the graph along the x-axis. 3) Additional terms like b and c shift the graph along the x or y-axis, and the vertex point becomes (-b/2a, c/a - b^2/4a^2) for the equation y = ax^2

1. Graphs of quadratic equations:
Graphs of quadratic equations would serve as primary visual aid for
understanding the nature of quadratic equations. Such as where the
value is maximum/minimum, where the graph touches or intercepts
the axes, whether it is symmetrical about a line or about a point, and
many other features are immediately visible in the graph. It may be
noted that different graphs stand for different equations, even the
same graph with position of axes or its orientation changed stand for
different equations.
In this lecture we will discuss only single valued quadratic functions
like 𝑦 = 𝑥2
, 𝑦 = 𝑎𝑥2
, 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 etc, for which more than one
values of x may correspond to one value of y, but not vice versa, more
than one value of y may correspond to one value of x, such as, 𝑥2
= 𝑦,
but not 𝑥 = 𝑎𝑦2
+ 𝑏𝑦 + 𝑐 etc., although those equations and graphs
are as important as former. We shall discuss them in the next lecture.
Graph of 𝒚 = 𝒙𝟐
This is the main equation or let us say, parent equation ; all other cases
such as 𝑦 = 𝑥2
, 𝑦 = 𝑎𝑥2
, or 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 etc. can be reduced to
this form. We would do it in the following paragraphs. Make a table
to draw the graph like this
Input x -3 -2 -1 0 1 2 3 4 5
Output y 9 4 1 0 1 4 9 16 25

2. If you employ geogebra, no table is necessary for graphing. This is
only to understand how to make a graph on a graph paper or make a
sketch on paper. Put x values in the equation and calculate y values.
In this graph of the function 𝑦 = 𝑥2
observe the following:
1. The graph passes though the origin, (0, 0); if x = 0, we get y =0.
2. The graph is symmetrical about Y-axis x = 0, This happens when
the value of y does not change while putting – x in place of x.
Mark the point (2,4) and (–2,4) ; the x values are negative of
each other but y values are same. Verify this also by putting
– x in place of x in the equation 𝑦 = 𝑥2
.
3. The graph is a parabola which opens upwards (concave
upwards) and goes to ∞ in both the branches in +ve y direction,
the branches are𝑦 = 𝑥2
and𝑦 = (−𝑥)2
, though they are the
same equation. The two branches are mirror images of each
other if the axis of symmetry is considered as a mirror and
minimum value of y is 0 in both the branches for x = 0 and this
point (0, 0) is called vertex point of the graph.

3. 4. The quadratic function has two coinciding roots x = 0 and x = 0
for y = 0. The graph touches both axes at the point (0, 0) instead
of intercepting it at different points. The two roots would have
been different in the latter case.
Graph of 𝒚 = −𝒙𝟐
Make a table to draw the graph like this
Input x -3 -2 -1 0 1 2 3 4 5
Output y -9 -4 -1 0 -1 -4 -9 -16 -25
In this graph of the function 𝑦 = −𝑥2
observe the following:
1. The graph passes though the origin, (0, 0), if x = 0, we get y =0.
2. The graph is symmetrical about Y-axis x = 0, This happens when
the value of y does not change while putting – x in place of x.
Mark the point (2, –4) and (– 2, – 4) ; the x values are negative
of each other by y values are same. Verify this also by putting
– x in place of x in the equation 𝑦 = 𝑥2
.

4. 3. The graph is a parabola which opens downwards (convex
upwards) and goes to - ∞ in both the branches in -ve y direction,
the branches are𝑦 = 𝑥2
and𝑦 = (−𝑥)2
, though they are the
same equation. The two branches are mirror images of each
other if the axis of symmetry is considered as a mirror and
minimum value of y is 0 in both the branches for x = 0 and this
point (0, 0) is called vertex point of the graph.
4. The quadratic function has two coinciding roots x = 0 and x = 0
for y = 0. The graph touches both axes at the point (0, 0) instead
of intercepting it at different points. Roots would have been
different in the latter case. How do you reconcile the facts that
the same equation represents two different graphs. In fact there
are not two graphs here, merely two branches of the same
graph, hence both represent the same equation.
5. This is obtained just by rotating the graph of the equation 𝒚 = 𝒙𝟐
by 180 degrees about the x-axis. More about rotations shall be
explained later.
It is important to understand and remember the above points before
proceeding further.
Graphs of 𝒚 = 𝒂𝒙𝟐
To understand the graph of the function 𝑦 = 𝑎𝑥2
we give below three
graphs of
𝑦 = 𝑓(𝑥) = 𝑥2
, 𝑦 = 𝑔(𝑥) = 2𝑥2
and 𝑦 = ℎ(𝑥) =
1
2
𝑥2

5. The curves (graphs) are colored in black, green and blue
respectively. Observe the points A(2,4), B(1,2) and C(2,2) on
the respective graphs. Verify that they satisfy the equations of
the respective graphs. We observe that the graph of function
𝑔(𝑥) = 2𝑥2
is narrower than the graph of 𝑓(𝑥) = 𝑥2
and the
graph of ℎ(𝑥) =
1
2
𝑥2
is wider than the graph of 𝑓(𝑥) = 𝑥2
.
We observe that the factor ‘a’ in the graph of 𝑦 = 𝑎𝑥2
tantamount to magnification by a factor
𝟏
√𝒂
of the graph of 𝑦 =
𝑎𝑥2
in the scale along the X-axis. (Stretching or compressing
across axis of symmetry). For if 𝑦 = 𝑎𝑥2
is transformed into
𝑦 = 𝑋2
in Xy axes instead of xy axes, where we have taken 𝑋2
for 𝑎𝑥2
, then 𝑋must be adjusted for the space available for
graphing √𝑎𝑥 and hence the scale must be
𝟏
√𝒂
times for that of
x. Note that the graph of 𝒚 = 𝒂𝒙𝟐
opens upwards if a>0 and vice
versa. Also the minimum value of y = 0 for a>0, and maximum value

6. of y = 0 is a<0. For a = 0, this is no more a quadratic equation.
Graphs of 𝒚 = 𝒂𝒙𝟐
+ 𝒄
As long as the equation is not changed, the graph is not
changed. So write this equation as 𝑦 − 𝑐 = 𝑎𝑥2
, or𝑌 = 𝑋2
,
if 𝑦 − 𝑐 = 𝑌 and 𝑎𝑥2
= 𝑋2
.Without looking at the graph, one could
tell that 𝑦 = 𝑐 for x = 0, as if the X-axis is shifted up by c units. In fact
if we put Y = y – c , the new x-axis is shifted up by c units in both cases
whether a>0 or a<0. The fact 𝑎𝑥2
= 𝑋2
tells us that the x-axis is
compressed by a factor of
1
√𝑎
resulting in X-axis, as previously
explained.
Look at the graph below. This is merely a print-screen copy. In
geogebra app. Just type 𝑦 = 𝑎𝑥2
+ 𝑐 and the graph is immediately
presented with two sliders, one for varying a and another for varying

7. c, or just copy the following link and paste it on your browser. By
moving the bold point on the sliders one can understand how the
graph changes when coefficients of the equation is changed.
https://www.geogebra.org/m/xphkhzhr
Note that the vertex of the graph of 𝑦 = 𝑎𝑥2
+ 𝑐 is at (0, c) instead of
at (0, 0) as in the graph of 𝑦 = 𝑎𝑥2
so minimum value of y is c when
a>0 and maximum value of y = c when a<0.
Graphs of 𝒚 = 𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄
Now it would be easy to graph 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 and interpret.
Complete the square and get,

8. 𝑦 = 𝑎 (𝑥2
+ 2
𝑏
2𝑎
𝑥 +
𝑏2
4𝑎2 + (
𝑐
𝑎
−
𝑏2
4𝑎2))
Or, 𝑦 = 𝑎 (𝑥 +
𝑏
2𝑎
)
2
+ (
4𝑎𝑐−𝑏2
4𝑎2 ), 𝑦 − 𝐶 = 𝑎𝑋2
,or, say,
𝑌 = 𝑎𝑋2
.
Where 𝑥 +
𝑏
2𝑎
= 𝑋 and 𝐶 =
𝑐
𝑎
−
𝑏2
4𝑎2, 𝑌 = 𝑦 − 𝐶
In total this is the same graph of 𝒚 = 𝒙𝟐
with a
compression factor in x-axis of
𝟏
√𝒂
, a shift in x-axis by −
𝒃
𝟐𝒂
and a shift in y-axis by C
Note that the origin (0,0) in old xy axes is shifted to
The point (−
𝑏
2𝑎
,
𝑐
𝑎
−
𝑏2
4𝑎2) which is the vertex of the graph;
minimum value of y is
𝑐
𝑎
−
𝑏2
4𝑎2 =
4𝑎𝑐−𝑏2
4𝑎2 if a>0 and maximum
value of y when a<0.
Note that the discriminant √𝑏
2
− 4𝑎𝑐 has a role to play in maximum
/ minimum problems. Details would be discussed in a later chapter.
This is here to understand the graph of = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 .
To get a dynamic view where one can vary a, b and c, visit below:
https://www.geogebra.org/classic/a7cxzksb

9. Vary the sliders to vary the parameters a, b and c.
A print screen image is also given below
To put a full stop to any confusion that might have been raised about
shifting of axes, let us observe what happens when the origin (0,0)
in old x-y axes is shifted to the point (h, k) and take this as origin of
the new axes of X and Y which are parallel to the old axes
respectively. The origin in the new X-Y axes shall be at (0, 0)
whereas this very point shall be (h, k) in the old axes. Thus any point
(A, B) on the new axes shall be (A + h, B + k) in the old axes.
Conversely any point (a, b) in the old axes shall be (a – h, b – k) in
the new axes. So no confusion now.
In the current example, the equation 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 which is
same as 𝑦 = 𝑎 (𝑥 +
𝑏
2𝑎
)
2
+ (
4𝑎𝑐−𝑏2
4𝑎2 ) may be written as

10. 𝑦 + (
𝑏2−4𝑎𝑐
4𝑎2 ) = 𝑎 (𝑥 +
𝑏
2𝑎
)
2
, or, 𝑌 = 𝑎𝑋2
,
where we have put 𝑦 + (
𝑏2−4𝑎𝑐
4𝑎2 ) = 𝑌 and 𝑥 +
𝑏
2𝑎
= 𝑋.
In effect, the origin is shifted to the point (−
𝑏
2𝑎
,− 𝑏
2
−4𝑎𝑐
4𝑎2 ).
This is the vertex and −
𝒃𝟐−𝟒𝒂𝒄
𝟒𝒂𝟐 =
𝟒𝒂𝒄−𝒃𝟐
𝟒𝒂𝟐 is maximum value of y
if a<0 and vice versa. The vertical line 𝒙 = −
𝒃
𝟐𝒂
is axis of symmetry.
Now the equation 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 is in the form
𝒚 + (
𝒃𝟐−𝟒𝒂𝒄
𝟒𝒂𝟐 ) = 𝒂 (𝒙 +
𝒃
𝟐𝒂
)
𝟐
…………………………(A)
Or , say, 𝒚 − 𝒌 = 𝒂(𝒙 − 𝒉)𝟐
…………………………. (B)
where ℎ = −
𝑏
2𝑎
and 𝑘 = −
𝑏2−4𝑎𝑐
4𝑎2 .
The point (𝒉, 𝒌) is its vertex , 𝑥 = −
𝑏
2𝑎
is the equation of its axis of
symmetry, and ‘a’ serves as magnification factor across the axis of
symmetry.
So this is one important form in which any quadratic equation such
as 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 can be written, of course, except those in the
form 𝑥 = 𝑎𝑦2
+ 𝑏𝑦 + 𝑐. Nothing wrong with the last equation,
except for the fact that here y is not a single valued function. It has
the very same nature when x and y are interchanged.
Vertex form
If we write 𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 in the form

11. 𝑦 = 𝑎(𝑥 − ℎ)2
+ 𝑘 = 𝑓(𝑥) and put x = h in this equation,
Then 𝑘 = 𝑓(ℎ).
So we can write 𝒚 = 𝒂(𝒙 − 𝒉)𝟐
+ 𝒇(𝒉)……………….(C)
Where ℎ = −
𝑏
2𝑎
and 𝑘 = 𝑓(ℎ)
The roots of the equations in the forms (A) and (B)
The roots of the equations are given by intercepts of the graph with
x-axis. This may be found by putting y = 0 in the equations.
Putting y = 0 in (A) we get, 𝑎 (𝑥 +
𝑏
2𝑎
)
2
=
𝑏2−4𝑎𝑐
4𝑎2 + 0
Or, (𝑥 +
𝑏
2𝑎
)
2
=
𝑏2−4𝑎𝑐
4𝑎2 , i.e., 𝑥 = −
𝑏
2𝑎
± √
𝑏2−4𝑎𝑐
4𝑎2 .
Or, 𝜶, 𝜷 = −
𝒃
𝟐𝒂
±
√𝒃𝟐−𝟒𝒂𝒄
𝟐𝒂
, as before.
Of course 𝑏2
− 4𝑎𝑐 need be positive for real intercepts.
Again, putting y = 0 in (B), 𝑎(𝑥 − ℎ)2
= −𝑘
Or, (𝑥 − ℎ)2
= −
𝑘
𝑎
, or, 𝑥 = ℎ ± √−
𝑘
𝑎
Or 𝜶, 𝜷 = 𝒉 ± √−
𝒌
𝒂
For real roots, k and a must be of opposite signs, otherwise there
shall be no intercepts with the x-axis.