Question

Draw the graphs of \[y = {x^2}\] and \[y = 2 + x\] solve the equation \[{x^2} - x - 2 = 0\]

Answer 1

Hint: The graph of a polynomial function is a smooth continuous curve or a line depending on the abscissa and ordinate variation.
Abscissa: In a system of coordinates the distance from a point to the vertical or y- axis, measured parallel to the horizontal or x-axis, the co-coordinate.
Ordinate: In a system of coordinates the distance from a point to the horizontal or x- axis, measured parallel to the vertical or y-axis, the y- coordinate.
i.e.\[\left( {x,y} \right) = \](abscissa, co-ordinate)
A graph can be obtained by putting different values of x and then obtaining different values of y.
Therefore,

Complete step by step answer:
Given \[y = {x^2}\]and \[y = 2 + x\]
For \[y = {x^2}\]
Put \[x = 0,y = {\left( 0 \right)^2} \Rightarrow y = 0\]
Put \[x = 1,y = {\left( 1 \right)^2} \Rightarrow y = 1\]
We have \[\left( {0,0} \right)and\left( {1,1} \right)\]
Put \[x = 2,y = {\left( 2 \right)^2} \Rightarrow y = 4\]
We have, \[\left( {2,4} \right)\]
Now plotting \[\left( {xy} \right)ie\left( {0,0} \right);\left( {1,1} \right);\left( {2,4} \right)\]in the graph for \[y = {x^2}\]

Hence, we see that \[y = {x^2}\]is a curve
For \[y = 2 + x\]
Put \[x = 0,y = 2 + 0 = 2,\] we have \[\left( {0,2} \right)\]
Put \[x = 1,y = 2 + 1 = 3\] we have \[\left( {1,3} \right)\]
Put \[x = 2,y = 2 + 2 = 4,\] we have \[\left( {2,4} \right)\]
Now plotting point \[\left( {x{\text{ }},{\text{ }}y} \right)\] i.e. \[\left( {0{\text{ }},2} \right)\]\[\left( {1,3} \right)\] \[\left( {2,4} \right)\] on the graph

Hence, we see that \[y = 2 + x\] is a straight line.
To solve \[{x^2} - x - 2 = 0\]
By middle term splitting, we have product of coefficient of \[{x^2}\] is \[1\] and constant \[ - 2\] is \[ - 2\] and we know sum of \[ - 2 + 1\] is \[ - 1\].
Hence, \[{x^2} - x - 2 = 0\]
\[ \Rightarrow {x^2} - 2x + x - 2 = 0\]
\[ \Rightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\] [Taking common]
\[ \Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\] [Taking \[ - 2\] common]
We have \[\left( {x - 2} \right) = 0\]and \[x + 1 = 0\]
     \[x = 2andx = - 1\]

Note:
We can also solve equations like \[a{x^2} + bx + c\]by middle term splitting or by hit and trial method.
We can find the number of points in a graph.
Remember that x, and y are positive in Ist quadrant only.