Chapter 9, Problem 1MCQ

To determine

To graph the given equation and mention its domain and range.

Explanation of Solution

Given information :

Given equation is $y=x^2+3x+1$ .

Graph :

  

Interpretation :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

1. Highest point (if $a<0$) and be called the maximum. Here, the graph opens downwards.
2. Lowest point (if $a>0$) and be called the minimum. Here, the graph opens upwards.

Here, a is greater than zero, hence the parabola opens upwards, and has a minimum value.

A parabola always points to infinity, either negative or positive.

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

For example, to compute the value of ‘y’ when $x=-1$ , substitute the value of ‘x’ in equation $y=x^2+3x+1$

On substituting the value of the ‘x’ as $x=-1$ , it is,

  $y={(-1)}^2+3(-1)+1$

On simplifying this,

  $=1-3+1$

Finally, the value of ‘y’ is

  $=-1$

Table of values of ‘x’ and ‘y’ .

x	y
-2	-1
-1.5	-1.25
-1	-1
0	1
1	5

Connect all the given points in a curved manner to draw the graph.

Since the parabola extends to infinity, the domain of this equation is $\left\{-\infty <x<\infty \right\}$ and the range is $\left\{y|y\ge -1.25\right\}$ . This is because the vertex of the parabola is minimum and at $-1.25$ .