given the graph of f(x)
using interval notation to show intervals of increasing and decreasing and postive and negative
If you know the invervals over which the graph f(x) is increasing or decreasing, it is pretty simple to put it into interval notation. The key is to concentrate on the X-coordinates for the numbers that you use to write the notation and the positon of the graph on the Y-axis to assess where the graph is increasing (the Ys are getting bigger), decreasing (the Ys are getting smaller), positive (the graph is above the X-axis), and negative (the graph is below the X-axis). For instance, if you have a parabola that is facing down with a vertex at (2,8), then the 2 is going to be what you use in the interval notation in terms of increasing or decreasing. 2 is called an inflection point because that is where the graph switching from going up to going down. So, the graph is increasing from negative infinity to 2 and decreasing from 2 to positive infinity. The interval notation would look like this: (-∞, 2) u (2,∞). Always use a parenthesis, not a bracket, with infinity or negative infinity. You also use parentheses for 2 because at 2, the graph is neither increasing or decreasing - it is completely flat. To find the intervals where the graph is negative or positive, look at the x-intercepts (also called zeros). Think of each intercept as a point on the number line. Any area between two intercepts that is below the X-axis is an interval where the function is negative. It that region of the graph is above the X-axis, it is positive. So, if our parabola facing down has x-intercepts at -3 and 7, the interval notation for where it is positive would read [-3,7]. You would use brackets because the graph is zero at -3 and zero at 7. Since zero is a positive number, these spots are included in the answer. In this case the middle hump of the graph is positive, but the ends regions on either side go below the X-axis. If the graph was facing up and the vertex was below the X-axis (like (2,-5)), the answer would read (-∞, -3] U (7, ∞). In this case, the two tips of the graph would extend above the X-axis, but the middle dip would be negative so it is the part not in the answer. If the graph floats completely above the X-axis, the interval for which it is positive would be all real numbers and the interval for which it is negative would be no solution. The opposite is true when the graph faces down and floats below the X-axis.