# Question Video: Sketching the Graph of a Quadratic Function Using a Table Mathematics

Sketch the graph of the quadratic function 𝑓(𝑥) = 𝑥² − 1 on the domain interval the closed interval [−2, 2] by completing the table.

02:54

### Video Transcript

Sketch the graph of the quadratic function 𝑓 of 𝑥 equals 𝑥 squared minus one on the domain interval the closed interval from negative two to two by completing the following table.

We’ve been instructed to complete the sketch by using this table. Our first step is then to plug in each of these five values for 𝑥 and find the output value. First, 𝑓 of negative two equals negative two squared minus one, which is three. And then 𝑓 of negative one equals negative one squared minus one, which is zero. 𝑓 of zero equals zero squared minus one, which is negative one. 𝑓 of one, one squared minus one equals zero. 𝑓 of two equals two squared minus one, which is three.

Now let’s clear some space and think about what graphing quadratic function is like. First of all, we recall that a quadratic function has the shape called a parabola that opens either upward or downward. For a quadratic in standard form 𝑓 of 𝑥 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, if 𝑎 is greater than zero for 𝑎 is positive, the parabola will open upward. If 𝑎 is negative, the parabola will open downward. In our case, 𝑓 of 𝑥 equals 𝑥 squared minus one. Our 𝑎-value is one; it’s greater than zero. So our sketch will open upward.

Looking at our table, we’re dealing with 𝑥-values from negative two to two. And we know that this function is on the closed interval negative two to two. Our 𝑓 of 𝑥 values range from negative one to three. A sensible scale might look something like this for the 𝑥- and 𝑦-axis. We’ll work on our sketch by first plotting the points from our table: negative two, three; negative one, zero; zero, negative one; one, zero; and two, three. As we’ve already recognized, this parabola opens upward. Between these points, we sketch the shape of a parabola. Notice here that the parabola is symmetrical about the point zero, negative one.

We can see that that is the vertex as there is a symmetry about that point, both in our table and on the graph. Note here that as this is a closed domain, negative two, three and two, three are endpoints of this function. We wouldn’t extend the graph further past negative two or two. To find this sketch, we plotted points from a table and then connected these points with a smooth curve, which gives us a sketch of the function 𝑓 of 𝑥 equals 𝑥 squared minus one on the domain interval negative two, two.