# 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.

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

### 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

### 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).

### 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

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