Increasing and Decreasing Intervals
For a quadratic function f(x) = ax², the increasing and decreasing intervals depend on the sign of a, with the vertex at x = 0 as the turning point — a key concept in enVision Algebra 1 Chapter 8 for Grade 11. When a > 0 (opens up): f(x) = 2x² decreases on (-∞, 0) and increases on (0, ∞). When a < 0 (opens down): f(x) = -½x² increases on (-∞, 0) and decreases on (0, ∞). The function always transitions at its vertex, and the intervals are described using interval notation with open parentheses because the vertex itself is neither increasing nor decreasing.
Key Concepts
For a quadratic function $f(x) = ax^2$: If $a 0$: decreasing on $( \infty, 0)$ and increasing on $(0, \infty)$ If $a < 0$: increasing on $( \infty, 0)$ and decreasing on $(0, \infty)$.
Common Questions
On what intervals does f(x) = 2x² increase and decrease?
f(x) = 2x² has a > 0, so it opens upward. It decreases on (-∞, 0) and increases on (0, ∞), with the vertex at the origin as the turning point.
On what intervals does f(x) = -½x² increase and decrease?
f(x) = -½x² has a < 0, so it opens downward. It increases on (-∞, 0) and decreases on (0, ∞).
Why are increasing/decreasing intervals written with open parentheses?
The vertex is neither increasing nor decreasing — it is a single turning point. Open parentheses exclude it from both intervals, accurately describing the direction of change.
How does the value of a affect the increasing/decreasing behavior?
The sign of a determines the direction (positive opens up, turning from decreasing to increasing; negative opens down, turning from increasing to decreasing). The magnitude of a affects steepness but not the intervals.
Does a quadratic like f(x) = x² + 3 have different increasing/decreasing intervals than f(x) = x²?
No. Both have the vertex at x = 0 (since there is no horizontal shift), so both decrease on (-∞, 0) and increase on (0, ∞). Vertical shifts do not move the vertex horizontally.