How to Draw Graph of Quadratic Polynomial
1. Write the given quadratic polynomial
2. Calculate the zeros of the polynomial, if exist, by putting y = 0 i.e.,
3. Calculate the points where the curve meets y-axis by putting x = 0.
4. Calculate
if D > 0, graph cuts x-axis at two points.
D = 0, graph touches x-axis at one point.
D < 0, graph is far away from x-axis.
5. Find
6. Make a table of selecting values of x and corresponding values of y, two to three values on left and two to three values on right of turning point
7. Draw a smooth curve through these points by free hand. The graph so obtained is called a parabola.
Note:
The graph of quadratic polynomial is a parabola.
If a is +ve, graph opens upward.
If a is –ve, graph opens downward.
If D > 0, parabola cuts x-axis at two points i.e. it has two zeros.
If D = 0, parabola touches x-axis at one point i.e. it has one zero.
If D < 0, parabola does not even touch x-axis at all i.e. it has no real zero.
Some Solved Examples based on Graph of Quadratic Polynomial
Question:
Draw the graph of the polynomial
Solution:
Let
Since the coefficient of
a = 1, b = 2, c = –3, D =
As D > 0, so parabola cuts x-axis at two points.
If
x = 1 or x = –3
It shows the graph of
Vertex =
Question:
Draw the graph of the polynomial
Solution:
Here a = –2, b = 4, c = –4.
As D < 0, so parabola does not cut x-axis.
Hence, the graph does not intersect x-axis. There is no zero
As coefficient of x2 is –ve, parabola opens downward.
Thus parabola intersect y-axis at (0, –4).
Vertex =
| x | –2 |
4x | |
| –1 | –2 | –4 | –10 |
| 0 | 0 | 0 | –4 |
| 1 | –2 | 4 | –2 |
| 2 | –8 | 8 | –4 |
| 3 | –18 | 12 | –10 |
Axis of symmetry : x = 1
Question:
Draw the graph of the polynomial
Solution:
So, it has three zeros.
Putting y = 0, we get x = 0, 2, –2
Hence, the zeroes of the polynomial are (0, 0), (2, 0) and (–2, 0)
Table for
| x | –4x | ||
| –3 | –27 | 12 | –15 |
| –2 | –8 | 8 | 0 |
| –1 | –1 | 4 | 3 |
| 0 | 0 | 0 | 0 |
| 1 | 1 | –4 | –3 |
| 2 | 8 | –8 | 0 |
| 3 | 27 | –12 | 15 |
By Plotting the ordered pairs (–3,–15),
Question:
Find zeros, if any and the vertex of the graph for the function
Solution:
Þ
a = 2, b = –4, c = 5
D =
= 16 – 40 = –24
As D < 0, so parabola does not cut x-axis at all, it means that it has no zero.
Hence the graph does not intersect x-axis.
As coefficient of x2 is +ve, parabola opens upward.
Putting x = 0, we get
The parabola intersect y-axis at (0, 5)
Vertex
= (1, 3)
Required table for
| x | 2 |
–4x | |
| –3 | 18 | 12 | 35 |
| –2 | 8 | 8 | 21 |
| –1 | 2 | 4 | 11 |
| 0 | 0 | 0 | 5 |
| 1 | 2 | –4 | 3 |
| 2 | 8 | –8 | 5 |
| 3 | 18 | –12 | 11 |
| 4 | 32 | –16 | 21 |
| 5 | 50 | –20 | 35 |
Zeros : Nil
Axis of symmetry : x = 1