## How to Draw Graph of Quadratic Polynomial

1. Write the given quadratic polynomial as

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 which is the turning point of curve.

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 is the given polynomial.

Since the coefficient of is positive, it will open upward.

a = 1, b = 2, c = –3, D = = 4 + 12 = 16

As D > 0, so parabola cuts x-axis at two points.

If Þ 0

x = 1 or x = –3

It shows the graph of will intersect x-axis at (1, 0) and (–3, 0).

Vertex = =

### Question:

Draw the graph of the polynomial .

### Solution:

is the given polynomial.

Here a = –2, b = 4, c = –4.

16 – 32 = –16

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 = = = (1, –2). Required table for

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 . Read off zeros from the graph.

### Solution:

is a cubic polynomial

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), , (0, 0), (1, –3), (2, 0) and on the graph paper. We find that the graph cuts x-axis at three points. So, it has three zeros, , . It has two vertices (1, –3) and (–1, 3).

### Question:

Find zeros, if any and the vertex of the graph for the function and also draw its graph. Also draw the axis of symmetry for this function.

### Solution:

if y = 0

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