There are no x terms on the right hand side so assume f(x)=-½x²+3.
The table of values needs to cover when f(x)=0, f(x)>0, f(x)<0 and the vertex.
Start with the vertex. f(x) has a maximum at f(x)=3 when x=0. Any other value of x produces a result less than 3. Also f(x) is symmetrical about the y axis (vertical axis) because f(x)=f(-x).
Now consider when the graph cuts the x axis (x intercepts), that is, when f(x)=0; -½x²+3=0 when x²=6 so x=±√6=±2.45 approx. The suggest range of values for x is -3 to 3. This covers all the essential points.
| x |
f(x) |
| -3 |
-1.5 |
| -2 |
1 |
| -1 |
2.5 |
| 0 |
3 |
| 1 |
2.5 |
| 2 |
1 |
| 3 |
-1.5 |
Vertex is shown as bold in the table and as a point on the graph.
