Question

Draw the graph of i) $y = 2x + 5$, ii) $y = 2x - 5$, iii) \[y = 2x\] and find the point of intersection on the $x$ axis. Is the $x$- coordinates of these points also the zero of the polynomial?

Answer 1

Hint: We will first find the set of points for the given equations by substituting the different values of $x$ and finding the values of $y$. We will then plot the points and draw a continuous line joining the points of every equation. For each equation, we will also find the point of intersection of the graph with the $x$ axis.

Complete step-by-step answer:
We will first let different values of $x$.
And find the values of $y$ for different values of $x$.
Let $y = 2x + 5$ (1)
Let $x = - 2$, then $y = 2\left( { - 2} \right) + 5 = 1$
Let $x = - 1$, then $y = 2\left( { - 1} \right) + 5 = 3$
Let $x = 0$, then $y = 2\left( 0 \right) + 5 = 5$
Let $x = 1$, then $y = 2\left( 1 \right) + 5 = 7$

$x$	$y$
$ - 2$	1
$ - 1$	3
0	5
1	7

We will now plot the points and then join the points with the line to plot the graph $y = 2x + 5$.

From the graph, we can see the point of intersection of the equation $y = 2x + 5$ with $x$ axis is $\left( { - \dfrac{5}{2},0} \right)$
Similarly, we will now find the values of $y$ for different values of $x$ for $y = 2x - 5$
Let $y = 2x - 5$ (2)
Let $x = - 2$, then $y = 2\left( { - 2} \right) - 5 = - 9$
Let $x = - 1$, then $y = 2\left( { - 1} \right) - 5 = - 7$
Let $x = 0$, then $y = 2\left( 0 \right) - 5 = - 5$
Let $x = 1$, then $y = 2\left( 1 \right) - 5 = - 3$

$x$	$y$
$ - 2$	$ - 9$
$ - 1$	$ - 7$
0	$ - 5$
1	$ - 3$

We will now plot the points and then join the points with the line to plot the graph $y = 2x - 5$.

From the graph of $y = 2x - 5$, we can see that the point of intersection of the graph with the $x$ axis is $\left( {\dfrac{5}{2},0} \right)$
Similarly, find the values of $y$ for different values of $x$ for \[y = 2x\]
Let \[y = 2x\] be equation (3)
That is, if $x = - 2$, then $y = 2\left( { - 2} \right) = - 4$
Let $x = - 1$, then $y = 2\left( { - 1} \right) = - 2$
Let $x = 0$, then $y = 2\left( 0 \right) = 0$
Let $x = 1$, then $y = 2\left( 1 \right) = 2$
Hence, write the values in the table for equation (3)

$x$	$y$
$ - 2$	$ - 4$
$ - 1$	$ - 2$
0	0
1	2

Now, we will plot the points and draw the corresponding graph of \[y = 2x\] with the other two graphs.

Here, the point of intersection of $y = 2x$ with the $x$ axis is $\left( {0,0} \right)$
Yes, the point of intersection of the graph with the $x$ axis is the zero of the given polynomial.
Note: The zero of the polynomial is the value of $x$ which gives the value of $y$ as zero. Here, we have calculated the point of intersection of the polynomial with the $x$ axis. The value of the $x$ coordinates of such points are the zeroes of the polynomial.