Examining the table, we see some clues that will help us write the function. We can, for example, identify both of the function's x-intercepts. These occur when y=0 and from the table we can see two of them.
x
-4
-3
-2
-1
0
1
2
3
4
5
6
7
y
18
8
0
-6
-10
-12
-12
-10
-6
0
8
18
As we can see, the function intercepts the x-axis at x= -2 and x= 5. With this information, we can begin writing our quadratic function in factored form.
y=a(x-( -2))(x- 5)
⇕
y=a(x+2)(x-5)
To complete the equation we have to find the value of a. To do that, we must substitute a third point that is not one of the x-intercepts into the equation and solve for a. Examining the table, we notice that we have also been given the function's y-intercept, which occurs at x=0.
x
-4
-3
-2
-1
0
1
2
3
4
5
6
7
y
18
8
0
-6
-10
-12
-12
-10
-6
0
8
18
By substituting (0,-10) into the function, we can solve for a. We could also use any of the other points but working with zeroes is generally easier.
Now we can complete the equation by substituting a= 1 and simplifying.
y= 1(x+2)(x-5)
⇕
y=(x+2)(x-5)
To check if the function is correct, we will plot it in a coordinate plane along with the points from the value table. If all points fall on the graph, we have the correct function.
