Question

Determine the consecutive integer values of x between which each real zero of (the equation in the photo) is located by using a table. Then sketch the graph.

Answer 1

In order to locate the zeros of the function, we need to make a table by giving integer values to the variable x.

For instance, if x=-4, we have

$\begin{gathered} f(-4)=(-4)^2+3(-4)-1 \\ f(-4)=16-12-1 \\ f(-4)=3 \end{gathered}$

Similarly, if x= -3, we get

$\begin{gathered} f(-3)=(-3)^2+3(-3)-1 \\ f(-3)=9-9-1 \\ f(-3)=-1 \end{gathered}$

When x= -2, we obtain

$\begin{gathered} f(-2)=(-2)^2+3(-2)-1 \\ f(-2)=4-6-1 \\ f(-2)=-3 \end{gathered}$

Now, when x= -1, we have

$\begin{gathered} f(-1)=(-1)^2+3(-1)-1 \\ f(-1)=1-3-1 \\ f(-1)=-3 \end{gathered}$

When x=0, we h ave

$\begin{gathered} f(0)=(0)^2+3(0)-1 \\ f(0)=-1 \end{gathered}$

and when x=1, we obtain

$\begin{gathered} f(1)=(1)^2+3(1)-1 \\ f(1)=1+3-1 \\ f(1)=3 \end{gathered}$

Then, by means of these results, the table is given by

From our table we can see that the graph change its sign between x=-4 and x=-3 and also when x=0 and x=1. This means that the graph crosses x-axis between these pair of numbers.

So we can conclude that there is a root (or zero) of the function between:

$\begin{gathered} x=-4\text{ and x=-3} \\ \text{and} \\ x=0\text{ and x= 1} \end{gathered}$

Let's corraborate this result by making the graph:

We can see that one zero ocurrs at x= -3.303 and the other at x= 0.303, which corroborate our results from above.