Graphing the Parabola y = ax² + bx + c

1.    Determine whether the parabola opens upward or downward.

a.    If a > 0, it opens upward.

b.    If a < 0, it opens downward.

2.    Determine the vertex.

a.    The x-coordinate is .

b.    The y-coordinate is found by substituting the x-coordinate, from          Step 2a, in the equation y = ax² + bx + c.

3.    Determine the y-intercept by setting x = 0.

4.    Determine the x-intercepts (if any) by setting y = 0, i.e., solving the equation
       ax² + bx + c = 0.

5.    Determine two or three other points if there are no x-intercepts.

 

EXAMPLES

Graph by determining all the key features of the graph. That is, find the vertex, the x- and y-intercepts (if any), and additional points if needed. Find the domain and range of the function.

Opens upward since the coefficient of the x² is positive

Vertex: and So the vertex is (3, -4).

y-intercept: The y-intercept is (0,5).

x-intercepts:

The x-intercepts are (5,0) and (1,0).

Points plotted on grid below:

The axis of symmetry is a vertical line passing through the vertex. Notice on the graph below that whatever is on one side of the axis of symmetry is also on the other side of the line.

The point (1, 0) is 2 units to the left of the axis of symmetry, and the point (5,0) is 2 units to the right of the axis of symmetry. So, since the point (0,5) is 3 units to the left of the axis of symmetry, the point (6,5) which is 3 units to the right of the axis of symmetry, is another point on the graph.

This gives us a fifth point for our graph.

CHECK: If x = 6, y = 5.

We could continue plotting points because the more points we plot, the more accurate our graph would be. However, that would take a lot of time. Plotting 5 points is sufficient - if one point is the vertex and there are two points on each side of the axis of symmetry.

Then connect the points with a smooth curve.

Look at the graph. Notice that the values of x continue in both directions. Therefore the domain is (-¥ ,¥ ) or all real numbers. As far as the range is concerned, the values of y begin at -4 and continue up. So the range is [-4,¥ ).

2. Graph

The graph of this parabola opens down because the leading coefficient is negative.

Vertex: and The vertex is the point (0,-3).

y-intercept: (0,-3)

x-intercepts:

There are no x-intercepts.

So after all of our steps, we have one point: (0,-3). We will have to make a table of values to complete our graph.

We can now use these points and the points symmetric to these to have our 5 points.

The point symmetric to (-1,-4) is (1,-4) and the point symmetric to (-2,-7) is (2,7). The axis of symmetry is the y-axis in this problem.

Connect the points with a smooth curve.

The domain of this function is again (-¥ ,¥ ) and the range is (-¥ ,-3].