# How To Draw Graph Of Quadratic Polynomials

## Polynomials of Class 10

A quadratic polynomial graph is a parabola. The roots of the quadratic figure are points where the parabola cuts the x axis i.e. points where the quadratic polynomial value is zero. There are certain steps you need to follow while making graph for quadratic polynomials:

Write the given quadratic polynomial f(x)= ax2 + bx+ c as Y = ax2 + bx + c

Calculate the zeros of the polynomial, if exist, by putting y = 0 i.e., ax2 + bx + c

Calculate the points where the curve meets y-axis by putting x = 0.

Calculate D = b2 – 4ac

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.

Find which is the turning point of curve.

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

Draw a smooth curve through these points by free hand. The graph so obtained is called a parabola.

Q1. Draw the graph of the polynomial f(x) = x2 – 2x – 8

Sol. Let y = x2 - 2x - 8.

The following table gives the values of y or f(x) for various values of x.

 x -4 -3 -2 -1 0 1 2 3 4 5 6 y = x2 – 2x - 8 16 7 0 -5 -8 -9 -8 -5 0 7 16

Let us plot the points (-4, 16), (-3, 7), (-2, 0), (-1, -5), (0, - 8), (1, - 9), (2, - 8), (3, - 5), (4, 0), (5, 7) and (6, 16) on a graphs paper and draw a smooth free hand curve passing through these points. The curve thus obtained represents the graphs of the polynomial f(x) = x2 - 2x - 8. This is called a parabola.

The lowest point P, called a minimum point, is the vertex of the parabola. Vertical line passing through P is called the axis of the parabola. Parabola is symmetric about the axis. So, it is also called the line of symmetry.

### Observations:

For the graphs of the polynomial f(x) = x2 - 2x - 8, following observations can be drawn:

• The coefficient of x2 in f(x) = x2 - 2x - 8 is 1 (a positive real number) and so the parabola opens upwards.
• D = b2 - 4ac = 4 + 32 = 36 > 0. So, the parabola cuts X-axis at two distinct points.
• On comparing the polynomial x2 - 2x - 8 with ax2 + bx + c, we get a = 1, b = –2 and c = –8.

The vertex of the parabola has coordinates (1, -9) i.e. , where D ≡ b2 - 4ac.

• The polynomial f(x) = x2 – 2x – 8 = (x – 4) (x + 2) is factorizable into two distinct linear factors (x – 4) and (x + 2). So, the parabola cuts X-axis at two distinct points (4, 0) and (–2, 0). The x-coordinates of these points are zeros of f(x).

For additional information related to the subject you can check the Maths Formula section.

## Graph of a Cubic Polynomial

Graphs of a cubic polynomial does not have a fixed standard shape. Cubic polynomial graphs will always cross X-axis at least once and at most thrice.

Q1. Draw the graphs of the polynomial f(x) = x3 - 4x.

Sol. Let y = f(x) or, y = x2 – 4x.

The values of y for variable value of x are listed in the following table :

 x –3 –2 –1 0 1 2 3 y = x3 – 4x –15 0 3 0 –3 0 15

Thus, the curve y = x3 – 4x passes through the points (–3, –15), (–2, 0), (–1, 3), (0 ,0), (1, –3), (2, 0), (3, 15), (4, 48) etc.

Plotting these points on a graph paper and drawing a free hand smooth curve through these points, we obtain the graph of the given polynomial as shown figure.

#### Observations:

For the graphs of the polynomial f(x) = x3 – 4x, following observations are as follows:

1. The polynomial f(x) = x3 - 4x = x(x2 – 4) = x(x – 2) (x + 2) is factorizable into three distinct linear factors. The curve y = f(x) also cuts X-axis at three distinct points.
2. We have, f(x) = x (x – 2) (x + 2). Therefore 0, 2 and -2 are three zeros of f(x). The curve y = f(x) cuts X-axis at three points O (0, 0), P(2, 0) and Q (-2, 0).

### Relationship Between Zeros and Coefficient of a Quadratic Polynomial

Let α and β be the zeros of a quadratic polynomial f(x) = ax2 + bx + c. By facto r theorem ( x - α) and (x - β) are the factors of f(x).

∴ f(x) = k ( x - α) (x - β) are the factors of f(x)

⇒ ax2 + bx + c = k{x2 - (α + β)x + αβ}

⇒ ax2 + bx + c = kx2 - k(α + β)x + kαβ

Comparing the coefficients of x2, x and constant terms on both sides, we get a = k, b = –k (α+ β) and kαβ

Hence,

Sum of the zeros

Product of the zeros

#### REMARKS:

If α and β are the zeros of a quadratic polynomial f(x). The , the polynomial f(x) is given by

f(x) = k{x2 - (α + β)x + αβ}

or f(x) = k{x2 - (Sum of the zeros) x + Product of the zeros}

Q1. Find a quadratic polynomial whose zeros are 5 +√2 and 5 - √2.

Sol. Given α = 5 +√2, β = 5 - √2

∴ f(x) = k{x2 - (α + β)x + αβ}

Here,

and

= 25 – 2 = 23

∴ f(x) = k {x2 – 10x + 23}, where, k is any non-zero real number.

### Relationship Between Zeros and Coefficient of a Cubic Polynomial

Let α,β,γ be the zeros of a cubic polynomial f(x) = ax3 + bx2 + cx + d, a ≠ 0 Then, by factor theorem,(x - α), (x - β) and (x - γ) are factors of f(x). Also, f(x) being a cubic polynomial cannot have more than three linear factors.

∴ f(x) = k (x - α)(x - β) (x - γ)

⇒ ax3 + bx2 + cx + d = k (x - α)(x - β) (x - γ)

⇒ ax3 + bx2 + cx + d = k{x3 }

⇒ ax3 + bx2 + cx + d = k x3 - k

Comparing the coefficients of x3, x2, x and constant terms on both sides, we get

a = k, b = –k

⇒ Sum of the zeros

⇒ Sum of the products of the zeros taken two at a time

⇒ Product of the zeros

Remarks :

Cubic polynomial having α,β and γ as its zeros is given by

∴ f(x) = k (x - α)(x - β) (x - γ)

f(x) = k {x3 - } where k is any non-zero real number.

Q1. Form a cubic polynomial with zeros α = 3, β = 2, γ = –1 .

Sol. α = 3, β = 2, γ = –1

Required polynomial = k (x - α)(x - β) (x - γ)

= k[(x - 3) (x - 2) (x + 1)]

= k[(x2 - 5x + 6)(x + 1)]

= k[x3 - 5x2 + 6x + x2 - 5x + 6]

= k[x3- 4x2 + x + 6]

where k is a non-zero constant.

Students can also access the class 10 Math Notes from here.