CBSE 10th Mathematics | How to Draw Graph of Quadratic Polynomial

How to Draw Graph of Quadratic Polynomial

1.    Write the given quadratic polynomial \displaystyle f(x)=a{{x}^{2}}+bx+c as

\displaystyle y=a{{x}^{2}}+bx+c

2.    Calculate the zeros of the polynomial, if exist, by putting y = 0 i.e., \displaystyle a{{x}^{2}}+bx+c=0

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

4.    Calculate \displaystyle D={{b}^{2}}-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.

5.    Find \displaystyle \left( -\frac{b}{2a},\,\,-\frac{D}{4a} \right) 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.

Graph of Quadratic Polynomial

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 \displaystyle f(x)={{x}^{2}}+2x-3.

Solution:

Let \displaystyle y={{x}^{2}}+2x-3 is the given polynomial.

Since the coefficient of \displaystyle {{x}^{2}} is positive, it will open upward.

a = 1, b = 2, c = –3, D = \displaystyle {{b}^{2}}-4ac={{(2)}^{2}}-4\times 1\times -3 = 4 + 12 = 16

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

If \displaystyle y=0    Þ    \displaystyle {{x}^{2}}+2x-3=0

\displaystyle {{x}^{2}}+3x-x-3=0

\displaystyle x(x+3)-1\,(x+3)=0

\displaystyle (x-1)(x+3)=0

x = 1 or x = –3

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

Vertex =\displaystyle \left( -\frac{b}{2a},\,\,-\frac{D}{4a} \right) = \displaystyle \left( -\frac{2}{2},\,\,-\frac{16}{4} \right)=(-1,\,\,-4)

Graph of Quadratic Polynomial

Graph of Quadratic Polynomial

Question:

Draw the graph of the polynomial \displaystyle -2{{x}^{2}}+4x-4.

Solution:

\displaystyle f(x)=y=-2{{x}^{2}}+4x-4 is the given polynomial.

Here a = –2, b = 4, c = –4.

\displaystyle D={{b}^{2}}-4ac

\displaystyle ={{(4)}^{2}}-4\times (-2)\times (-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.

\displaystyle f(0)=0-4=-4

Thus parabola intersect y-axis at (0, –4).

Vertex = \displaystyle \left( -\frac{b}{2a},\,\,-\frac{D}{4a} \right)                         = \displaystyle \left( -\frac{4}{2\times (-2)},\,\,\frac{16}{4\times (-2)} \right) = (1, –2).         Required table for

x –2\displaystyle {{x}^{2}} 4x \displaystyle y=-2{{x}^{2}}+4x-4
–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

Graph of Quadratic Polynomial

Question:

Draw the graph of the polynomial \displaystyle y={{x}^{3}}-4x. Read off zeros from the graph.

Solution:

\displaystyle y={{x}^{3}}-4x 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 \displaystyle f(x)=\displaystyle {{x}^{3}}-4x

x \displaystyle {{x}^{3}} –4x \displaystyle y={{x}^{3}}-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), \displaystyle (-2,\,0),\,\,(-1,\,\,3), (0, 0), (1, –3), (2, 0) and \displaystyle (3,\,15) on the graph paper. We find that the graph cuts x-axis at three points. So, it has three zeros, \displaystyle A(-2,\,0),\,\,O(0,\,\,0), \displaystyle A\prime (2,\,\,0). It has two vertices (1, –3) and (–1, 3).

Graph of Quadratic Polynomial

Question:

Find zeros, if any and the vertex of the graph for the function \displaystyle y=2{{x}^{2}}-4x+5 and also draw its graph. Also draw the axis of symmetry for this function.

Solution:

\displaystyle y=2{{x}^{2}}-4x+5 if y = 0

Þ    \displaystyle 2{{x}^{2}}-4x+5=0

a = 2, b = –4, c = 5

D = \displaystyle {{b}^{2}}-4ac={{(-4)}^{2}}-4\times 2\times 5

= 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 \displaystyle f(0)=0+5=5

The parabola intersect y-axis at (0, 5)

Vertex \displaystyle \left( -\frac{b}{2a},\,\,-\frac{D}{4a} \right)=\left( \frac{4}{2\times 2},\,\,\frac{24}{8} \right)

= (1, 3)

Required table for \displaystyle y=2{{x}^{2}}-4x+5

x 2\displaystyle {{x}^{2}} –4x \displaystyle y=2{{x}^{2}}-4x+5
–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

Graph of Quadratic Polynomial

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