Graphical Representation of Quadratic Expressions and Solutions to Quadratic Equations
You have learned how to solve simultaneous linear equations through graphs. We will now learn how to draw graphs of quadratic expressions in one valuable and, hence, solve to find the maxima, minima and zeros of the quadratic equation from the graph.
Consider these examples.
Example 1
Draw the graphs of the following functions.
 f(x)=x^{2}
 f(x)=2x^{2}
a. Let y=f(x)=x^{2}; we first obtain the points to be plotted
x 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
(3)^{2} 
(2)^{2} 
(1)^{2} 
0 
12 
22 
32 
Y 
9 
4 
1 
0 
1 
4 
9 
Plot the points (3,9), (2,4), (1,1), (0,0), (1,1), (2,4), (3,9) and join them to form a smooth curve. The curve formed is called a parabola.
b. Let y=f(x)=2x^{2}
x 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
(3)^{2} 
(2)^{2} 
(1)^{2} 
0 
(1)^{2} 
(2)^{2} 
(3)^{2} 
2x^{2} 
29 
24 
21 
0 
21 
24 
29 
y 
18 
8 
2 
0 
2 
8 
18 
Plot the points (3,18), (2,8), (1,2), (0,0), (1,2), (2,8), (3,18) and join them to form a smooth curve. Both graphs are parabolas.
As both pass through the point (0,0), they have common vertex. They also lie above the xaxis and are symmetric about the yaxis.
Example 2
Draw the graph of
x 
4 
2 
0 
2 
4 
x^{2} 
(4)^{2} 
(2)^{2} 
0 
(2)^{2} 
(4)^{2} 



0 


y 
8 
2 
0 
2 
8 
Plot the points (4,8). (2,2), (0,0), (2,2), (4,8) and join them to form a smooth curve.
Look at the graph of y = 1/2 x^{2}. It is also a parabola that lies below the xaxis in the third and fourth quadrants and is symmetrical about the yaxis.
Example 3
Draw the graph for y = 3x^{2}
x 
2 
1 
0 
1 
2 
x^{2} 
(2)^{2} 
(1)^{2} 
0 
(1)^{2} 
(2)^{2} 
3x^{2} 
34 
31 
0 
31 
34 
y 
12 
3 
0 
3 
12 
Example 4
Draw the following curves and make a comparison.
 f(y) = y^{2}
 f(y) = 2y^{2}
 f(y) = 1 / 3 y^{2}
 f(x) = y^{2}
a. Let x = f(y) = y^{2}
We can find the points to be plotted as follows.
y 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
(3)^{2} 
(2)^{2} 
(1)^{2} 
0 
12 
22 
32 
x 
9 
4 
1 
0 
1 
4 
9 
However, when we plot the points, we take them in the form (x,y).
Plot the points (9,3), (4,2), (1,1), (0,0), (1,1), (4,2), (9,3) and join them to form a smooth curve.
b. Let x = f(y) = 2y^{2}
y 
3 
2 
1 
0 
1 
2 
3 
y^{2} 
(3)^{2} 
(2)^{2} 
(1)^{2} 
0 
12 
22 
32 
2y^{2} 
2*9 
2*4 
2*1 
0 
2*1 
2*4 
2*9 
x 
18 
8 
2 
0 
2 
8 
18 
Plot the points (18,3), (8,2), (2,1), (0,0), (2,1), (8,2), (18,3) and join them to form a smooth curve.
c. Let x = 1/3 y^{2}
y 
6 
3 
0 
3 
6 
y^{2} 
(6)^{2} 
(3)^{2} 
0 
32 
62 
1/3y^{2} 
1/3 * 36 
1/3 * 9 
0 
1/3 * 9 
1/3 * 36 
x 
12 
3 
0 
3 
12 
Plot the points (12,6), (3,3), (0,0), (3,3), (12,6) and join them to form a smooth curve.
d. Let x=y^{2}
y 
3 
2 
1 
0 
1 
2 
3 
y^{2} 
(3)^{2} 
(2)^{2} 
(1)^{2} 
0 
(1)^{2} 
(2)^{2} 
(3)^{2} 
x 
9 
4 
1 
0 
1 
4 
9 
Plot the points (9,3), (4,2), (1,1), (0,0), (1,1), (4,2), (9,3) and join them to form a smooth curve.
From the graphs, we find that the curves x = y^{2}, x = 2y^{2} are parabolas symmetric about the xaxis and lie toward the right of the yaxis. They also lie in the first and fourth quadrants.
The curves x = 1/3 y^{2}, x = y^{2} are parabolas symmetric about the xaxis but lie to the left of the yaxis. They lie in the second and third quadrants.
Example 5
Draw the graph for f(x)=x^{2}x12
Let y=x^{2}x12
To plot the points, we find the value of y as follows.
f(3)=(3)^{2}(3)12
=9+312
=0
f(2) =(2)^{2}(2)12
=4212
=414
=10
In general, we use the following format.
y 
4 
3 
2 
1 
0 
1 
2 
3 
4 
5 
x^{2} 
16 
9 
4 
1 
0 
1 
4 
9 
16 
25 
x 
+4 
3 
2 
+1 
0 
1 
2 
3 
4 
5 
12 
12 
12 
12 
12 
12 
12 
12 
12 
12 
12 
y 
8 
0 
6 
10 
12 
12 
10 
6 
0 
8 
We plot the points (4,8), (3,0), (2,6), (1,10), (0,12), (1,12), (2,10), (3,6), (4,0), (5,8), and join them to form a smooth curve.
Observe that when y=0, x= 3 or 4 or y = 0 ⇒ x^{2}x12=0
∴ the roots or solutions of x^{2}x12=0 are x=3, 4
Further, the curve cuts the xaxis at the points (3,0) and (4,0). So the roots of the equation are the xcoordinates of the points, where the curve cuts the xaxis.
By looking at the graph, we can solve x^{2}x12=0 by taking the points where the curve cuts the xaxis, i.e., (3.0), (4,0).
We will now consider the graphs for equations of the form y=ax^{2}+bx+c and see what happens when we change the coefficients of the x^{2} and x terms, and the constant term.
First, we consider the quadratic equations of the form
y = ax^{2}+b+c where the coefficient of x^{2}, which is a, changes.
Example 6
Draw the graphs of the following quadratic equations.
 y = x^{2}3x1
 y = 2x^{2}3x1
 y = 3x^{2}3x1
 y = x^{2}3x1
i. y = x^{2}3x1
x 
2 
1 
0 
1 
2 
3 
4 
x^{2} 
4 
1 
0 
1 
4 
9 
16 
3x 
+6 
+3 
0 
3 
6 
9 
12 
1 
1 
1 
1 
1 
1 
1 
1 
y 
9 
3 
1 
3 
3 
1 
3 
Plot the points (2,9), (1,3), (0,1), (1.3), (2,3), (3,1), (4,3) and join them to form a smooth curve.
ii. y = 2x^{2}3x1
x 
1 
0 
1 
2 
3 
x^{2} 
1 
0 
1 
4 
9 
2x^{2} 
2 
0 
2 
8 
18 
3x 
3 
0 
3 
6 
9 
1 
1 
1 
1 
1 
1 
y 
4 
1 
2 
1 
8 
Plot the points (1,4), (0,1), (1,2), (2,1), (3,8) and join them to form a smooth curve.
iii. y=3x^{2}3x1
x 
1 
0 
1 
2 
x^{2} 
1 
0 
1 
4 
3x^{2} 
3 
0 
3 
12 
3x 
3 
0 
3 
6 
1 
1 
1 
1 
1 
y 
5 
1 
1 
5 
Plot the points (1,5), (0,1), (11), (2,5) and join them to form a smooth curve.
iv. y=x23x1
X 
4 
3 
2 
1 
0 
1 
2 
x^{2} 
16 
9 
4 
1 
0 
1 
4 
3x 
12 
+9 
+6 
+3 
0 
3 
6 
1 
1 
1 
1 
1 
1 
1 
1 
Y 
5 
1 
1 
1 
1 
5 
11 
Plot the points (4,5), (3,1), (2,1), (1,1), (0,1), (1,5), (2,11) and join them to form a smooth curve.
The scale taken for the graph is
xaxis 1 unit = 2 cm
yaxis 1 unit = 1cm
On the xaxis, each decimal fraction is denoted by two divisions so that 0.2 = 2* 2 = 4 small divisions on the graph.
1.2 = 1.2*2 = 24 small divisions on the graph.
Comparing the graphs, we notice that the curves i, ii, iii, iv all meet at the point (0,1)
All of the curves cross the xaxis so their roots are real.
Curve i: y = x^{2}3x1 meets the xaxis at the points
(0.3,0) and (3.3,0)
So the roots of x^{2}3x1 = 0 are x = 0.3, 3.3
Curve ii: y = 2x^{2}3x1 meets the xaxis at the points
(0.3,0) and (1.7,0)
So the roots of 2x^{2}3x1 = 0 are x = 0.3, 1.7
Observe that curves i and ii intersect
at (0,1) and (0.3,0)
Curve iii: y = 3x^{2}3x1 meets the xaxis at (0.2,0), (1.2,0)
So the roots of 3x^{2}3x1 = 0 are x = 0.2, 1.2
Curve iv y = x^{2}3x1 meets the xaxis at the points
(2.6,0) and (0.4,0)
So the roots of the equation x^{2}3x1 = 0 are
x = 2.6, 0.4
Now consider the quadratic equations where
the coefficient of x, b, changes.
Example 7
Consider the following quadratic equations.
 y = x^{2}5x+1
 y = x^{2}3x+1
 y = x^{2}+2x+1
 y = x^{2+4x+1}
i. y = x^{2}5x+1
x 
2 
1 
0 
1 
2 
3 
4 
x^{2} 
4 
1 
0 
1 
4 
9 
16 
5x 
+10 
+5 
0 
5 
10 
15 
20 
+1 
+1 
+1 
1 
1 
1 
1 
1 
y 
15 
7 
1 
3 
5 
5 
3 
Plot the points (2,15), (1,7), (0,1), (1,3), (2,5), (3,5), (4,3) and join them to form a smooth curve.
ii. y = x^{2}3x+1
x 
2 
1 
0 
1 
2 
3 
x^{2} 
4 
1 
0 
1 
4 
9 
3x 
6 
3 
0 
3 
6 
9 
1 
1 
1 
1 
1 
1 
1 
y 
11 
5 
1 
1 
1 
1 
Plot the points (2,11), (1,5), (0,1), (1,1), (2,1), (3,1) and join them to form a smooth curve.
iii. y = x^{2}+2x+1
x 
3 
2 
1 
0 
1 
2 
x^{2} 
9 
4 
1 
0 
1 
4 
2x 
6 
4 
2 
0 
2 
4 
+1 
+1 
1 
1 
1 
1 
1 
y 
4 
1 
0 
1 
4 
9 
Plot the points (3,4), (2,1), (1,0), (0,1), (1,4), (2,9) and join them to form a smooth curve.
iv. y = x^{2}+4x+1
x 
4 
3 
2 
1 
0 
1 
x^{2} 
16 
9 
4 
1 
0 
1 
4x 
16 
12 
8 
4 
0 
4 
1 
1 
1 
1 
1 
1 
1 
y 
1 
2 
3 
2 
1 
6 
Plot the points (4,1), (3,2), (2,3), (1,2), (0,1), (1,6) Scale.
xaxis 1unit = 2 cm
yaxis 1unit = 1 cm
We can make the following observations.
All pf the curves intersect at the point (0,1).
Curve i: y=x^{2}5x+1 meets the xaxis at the points
(0.2,0), and (4.8,0)
So the roots of x^{2}5x+1=0 are x=0.2, x=4.8
Curve ii: y=x^{2}3x+1 meets the xaxis at the points
(0.4,0) and (2.6,0)
So the roots of x^{2}3x+1=0 are x=0.4, 2.6
Curve iii: y=x^{2}+2x+1 meets the xaxis at the point
(1,0). So the roots are repeated
So roots of x^{2}+2x+1=0 are x=1, 1
Curve iv: y=x^{2}+4x+1 meets the xaxis at (3.7,0)
and (0.3,0)
So roots of x^{2}+4x+1=0 are x=3.7, 0.3
We now consider the curve where the constant term c of ax^{2}+bx+c varies.
Example 8
Consider the following quadratic equations.
 y = x^{2}+2x3
 y = x^{2}+2x+2
 y = x^{2}+2x6
 y = x^{2}+2x+4
i. y = x^{2}+2x3
x 
4 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
16 
9 
4 
1 
0 
1 
4 
9 
2x 
8 
6 
4 
2 
0 
2 
4 
6 
3 
3 
3 
3 
3 
3 
3 
3 
3 
y 
5 
0 
3 
4 
3 
0 
5 
12 
Plot the points (4,5), (3,0), (2,3), (1,4), (0,3), (1,0), (2,5), (3,12) and join them to form a smooth curve.
ii. y = x^{2}+2x+2
x 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
9 
4 
1 
0 
1 
4 
9 
2x 
6 
4 
2 
0 
2 
4 
6 
2 
2 
2 
2 
2 
2 
2 
2 
y 
5 
2 
1 
2 
5 
10 
17 
Plot the point (3,5), (2,2), (1,1), (0,2), (1,5), (2,10), (3,17) and join them to form a smooth curve.
iii. y = x^{2}+2x6
x 
4 
3 
2 
1 
0 
1 
2 
3 
4 
x^{2} 
16 
9 
4 
1 
0 
1 
4 
9 
16 
2x 
8 
6 
4 
2 
0 
2 
4 
6 
8 
6 
6 
6 
6 
6 
6 
6 
6 
6 
6 
y 
2 
3 
6 
7 
6 
3 
2 
9 
18 
Plot the points (4,2), (3,3), (2,6), (1,7), (0,6), (1,3), (2,2), (3,9), (4,18) and join them to form a smooth curve.
iv. y = x^{2}+2x+4
x 
2 
1 
0 
1 
2 
x^{2} 
4 
1 
0 
1 
4 
2x 
4 
2 
0 
2 
4 
4 
4 
4 
4 
4 
4 
y 
4 
3 
4 
7 
12 
Plot the points (2,4), (1,3), (0,4), (1,7), (2,12) and join them to form a smooth curve.
The scale is the same as the previous two examples.
We observe that, none of the curves intersect.
Curve i: y^{2}= x^{2}+2x3 intersects the xaxis at the points
(3,0), and (1,0)
The roots of x^{2}+2x3 = 0 are therefore x = 3,1
Curve iii: y = x^{2}+2x6 intersects the xaxis at the points
(3.6,0) and (1.6,0)
So the roots of x^{2}+2x6 = 0 are x = 3.6, 1.6
Curves ii and iv: that is y = x2+2x+2 and y = x2+2x+4,
do not touch or cut the xaxis.
So the quadratic equations x^{2}+2x+2 = 0 and x^{2}+2x+4 = 0 have no real roots. Their roots are complex and are obtained algebraically by using the formula x^{2}+2x+2 = 0
a = 1, b = 2, c = 2,
Discriminant = b24ac
= (2)2  4*1*2
= 4  8
= 4 < 0
Using the formula
x = 1+ i, 1i
x^{2}+2x+4 = 0
a = 1, b = 2, c = 4
Discriminant Δ = b24ac
= (2)^{2}4*1*4
= 416
= 12 < 0
Using the formula
Using the formula
From the graph, how does one obtain the roots?
Consider Curve iii. The curve cuts the xaxis at a distance of 7.2 cm from 0 to its left.
Since each unit = 2 cm, we divide 7.2 by 2
7.2/2 = 3.6
∴ The point is 3.6 units along the xaxis, which is the point (3.6,0)
The curve also cuts the xaxis at a distance of 3.2 cm from 0 to its right.
We therefore obtain 3.2/2 = 1.6 units.
So the point is actually (1.6, 0)
Example 9
Consider the following quadratic equations.
Draw the graphs of:
 y = 2x^{2}+x1
 y = x^{2}6x+9
 y = x^{2}4x+5
i. y = 2x2+x1
x 
3 
2 
1 
0 
1 
2 
x^{2} 
9 
4 
1 
0 
1 
4 
2x^{2} 
18 
8 
2 
0 
2 
8 
x 
3 
2 
1 
0 
1 
2 
1 
1 
1 
1 
1 
1 
1 
y 
14 
5 
0 
1 
2 
9 
Plot the points (3,14), (2,5), (1,0), (0,1), (1,2), (2,9) and join them to form a smooth curve.
ii. y = x26x+9
x 
1 
0 
1 
2 
3 
4 
x^{2} 
1 
0 
1 
4 
9 
16 
6x 
+6 
0 
6 
12 
18 
24 
9 
9 
9 
9 
9 
9 
9 
y 
16 
9 
4 
+1 
0 
1 
Plot the points (1,16), (0,9), (1,4), (2,1), (3,0), (4,1) and join them to form a smooth curve.
iii. y = x^{2}4x+5
x 
2 
1 
0 
1 
2 
3 
x^{2} 
4 
1 
0 
1 
4 
9 
4x 
8 
4 
0 
4 
8 
12 
5 
5 
5 
5 
5 
5 
5 
y 
17 
10 
5 
2 
1 
2 
Plot the points (2,17), (1,10), (0,5), (1,2), (2,1) and (3,2) and join them to form a smooth curve.
Curve i: y=2x^{2}+x1 cuts the xaxis at two points (1,0) and (+1 / 2), 0.
So 2x^{2}+x1=0 has 2 real roots x = 1, +1/2
Curve ii: y = x^{2}6x+9 touches the xaxis at the point (3,0).
The equation x^{2}6x+9 = 0 has two equal roots, or repeated roots, which are x = 3, 3.
Curve iii: y = x24x+5 does not touch or cut the xaxis at all. This means that x24x+5 = 0 has no real roots, but has only complex roots. These roots
have to be obtained algebraically.
x^{2}4x+5=0
a = 1, b = 4, c = 5.
Discriminant Δ = b24ac
= (4)^{2}  4*1*5
= 16  20
= 4 < 0
Using the formula, we get the roots as follows.
From this example, we infer the following about the roots of the equation.
ax^{2}+bx+c = 0 a≠ 0. a, b, c are real numbers (1)
 If the curve y = ax^{2}+bx+c meets the xaxis at two points (x1,0), (x2,0) then x1, x2 are two distinct real roots of equation (1)
 If the curve y = ax^{2}+bx+c meets the xaxis in exactly one point (x1,0), equation (1) has exactly one root, which is a coincident or repeated root.
 If the curve y = ax^{2}+bx+c never meets the xaxis, then its roots are complex numbers.
There is another way to obtain the roots of the equation.
ax^{2}+bx+c = 0 (1)
This can be rewritten as
ax^{2} = bxc (2)
Let y = ax2 (3)
and y = bxc (4)
Equation (3) forms a parabola.
Equation (4) is a straight line.
We next draw the graphs to the same scale.
 If in the graph, (4) cuts (3); that is, the straight line cuts the curve in two points so that their xcoordinates are the same. Then (1) has two real roots.
 If the line and the curve have a single point in common, then their xcoordinate is the only root of equation (1); that is, the root is repeated.
 If the line and the curve have no points in common, then equation (1) has only complex roots.
The following example will illustrate these points.
Example 10
Solve graphically
x^{2}+2x15 = 0
This can be rewritten as
x^{2} = 2x+15
Let y = x^{2}
and y = 2x15
y = x^{2}
x 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
9 
4 
1 
0 
1 
4 
9 
4y 
9 
4 
1 
0 
1 
4 
9 
Plot the points (3,9), (2,4), (1,1), (0,0), (1,1), (2,4), (3,9) and join them to form a smooth curve.
Y = 2x+15
x 
5 
4 
3 
2 
1 
0 
1 
2 
3 
2x 
10 
8 
6 
4 
2 
0 
2 
4 
6 
15 
15 
15 
15 
15 
15 
15 
15 
15 
15 
y 
25 
23 
21 
19 
17 
15 
13 
11 
9 
Plot the points and join them (5,25), (4,23), (3,21), (2,19), (1,17), (0,15), (1,13), (2,11), (3,9).
From the graph, y = 2x+15 meets y = x^{2} at the points (5,25) and (3,9) since the xcoordinates represent the roots of the equation.
The roots of x^{2}+2x15=0 are x=5, 3
Try these questions
Solve graphically
 4x^{2}  4x+1 = 0
This can be rewritten as
4x^{2}=4x1
Let y = 4x^{2}
and y = 4x1
y = 4x^{2}
X 
3 
2 
1 
0 
1 
2 
3 
x^{2} 
9 
4 
1 
0 
1 
4 
9 
4x^{2} 
36 
16 
4 
0 
4 
16 
36 
Y 
36 
16 
4 
0 
4 
16 
36 
Plot the points (3,36), (2,16), (1,4), (0,0), (1,4), (2,16), (3,36) and join them the form a smooth curve.
y = 4x1
x 
1 
4 
4x 
4 
16 
1 
1 
1 
y 
5 
15 
Plot the points (1,5), (4,15) and join them to form a line.
From the graph, the curve and line meet at exactly one point, (1/2,1).
So the roots of 4x^{2}4x+1=0 are repeated.
 x^{2}6x+10 = 0
This can be rewritten as
x^{2} = 6x10
Let y = x^{2}
and y = 6x10
y = x^{2}
x 
5 
4 
3 
2 
1 
0 
1 
2 
3 
4 
5 
x^{2} 
25 
16 
9 
4 
1 
0 
1 
4 
9 
16 
25 
y 
25 
16 
9 
4 
1 
0 
1 
4 
9 
16 
25 
Plot the points (5,25), (4,16), (3,9), (2,4), (1,1), (0,0), (1,1), (2,4), (3,9), (4,16), (5,25) and join them to form a smooth curve.
x 
0 
2 
3 
6x 
0 
12 
18 
10 
10 
10 
10 
y 
10 
2 
8 
Plot the points (0,10), (2,2), (3,8) and join them to from a straight line.
Since the line and curve do not intersect, the roots are complex.
Solving algebraically x^{2}6x+10 = 0
a =1, b = 6, c = 10