What happens if we change “a” in f(x) = a(x – h)² + k?

The parameter “a” has two possible effects on the graph:

1. If “a” is positive (a>0), then the parabola opens up.

If “a” is negative (a<0), the parabola reflects across the x-axis and opens down.

2. If |a| > 1, then the parabola is narrower than the graph of the parent function. (The parabola has been stretched.)

If 0<|a|<1, then the parabola is wider than the graph of the parent function. (The parabola has been compressed.)

• We use the absolute value of “a” because the sign of “a” does NOT affect the openness of the parabola.
• The sign of “a” ONLY determines whether the parabola opens up or down.

Example

Determine which of the functions is the widest: y = 1 over 2 1 2 x² or y = −3x².

Solution

The coefficient of x², “a”, determines . . .

• whether the parabola opens up or down and
• the openness of the parabola

For y = 1 over 2 1 2 x² ⇒ a = 1 over 2 1 2 .

For y = −3x² ⇒ a = −3.

• Remember that the sign of “a” ONLY determines whether the parabola opens up or down. The openness of a parabola is determined by |a|.

| 1 over 2 1 2 | = 1 over 2 1 2 and |−3| = 3

Since 1 over 2 1 2 < 3, then y = 1 over 2 1 2 x² is wider than y = -3x²

Example

Given the parabola f(x) = - 1 over 2 1 2 x². How would the graph be affected if - 1 over 2 1 2 is changed to 2?

Solution

The coefficient of x² , “a”, determines . . .

• whether the parabola opens up or down and
• the openness of the parabola

So, “a” is changing from − 1 over 2 1 2 to 2.

Since “a” changes from negative to positive, the graph reflects across the x-axis.

| 1 over 2 1 2 | = 1 over 2 1 2 and |−2| =2. Since 2 > 1 over 2 1 2 then the graph has stretched vertically and is now narrower.

Let's see this in action.