Make a table of values and connect the points with a smooth curve.
Table:
x
-4
-3
-2
-1
0
1
2
y
5
0
-3
-4
-3
0
5
Graph:
Roots: (-3,0) and (1,0)
Practice makes perfect
We want to complete the given table for the equation y = x^2+2x-3. Let's substitute the given x-values into the equation and simplify, starting with x = -4.
We found that y=5 when x = -4. Let's repeat this process for other inputs from the table.
x
x^2+2x-3
y = x^2+2x-3
-4
( -4)^2+2( -4)-3
5
-3
( -3)^2+2( -3)-3
0
-2
( -2)^2+2( -2)-3
-3
-1
( -1)^2+2( -1)-3
-4
0
0^2+2( 0)-3
-3
1
1^2+2( 1)-3
0
2
2^2+2( 2)-3
5
Let's complete the table using the y-values we found.
x
-4
-3
-2
-1
0
1
2
y
5
0
-3
-4
-3
0
5
Notice that we found two roots, or x-intercepts, in our table — the points (-3,0) and (1,0). We know these are the roots of the equation because the y-values are 0, so the graph crosses the x-axis at these points. Now let's plot the points from the table and connect them with a smooth curve.
