How am I supposed to graph g(x) = |x|/2 - 3/4?
Make a t chart
and plug in an initial x to get your y to get your points then graph.
for example:
pick 1 for x as your starting
y=|1|/2 - 3/4
y=-1/4
so first points are
x=1
y=-1/4
x=2
y=???
x=3
y=???
so on..
should get a diagonal line
To graph the function g(x) = |x|/2 - 3/4, you can follow these steps:
1. Find the key points: Identify the points where the function g(x) may change direction or cross the x-axis. Since g(x) has an absolute value function, its key points are when the expression inside the absolute value sign is zero. In this case, set |x|/2 - 3/4 equal to zero and solve for x: |x|/2 - 3/4 = 0.
Solving for x, you get:
|x|/2 = 3/4
|x| = 6/4
Since the absolute value of any number is always positive, x can be either 6/4 or -6/4, which simplifies to 3/2 or -3/2 respectively. Therefore, the key points are x = 3/2 and x = -3/2.
2. Determine the behavior of the function at the key points:
- When x < -3/2, the expression inside the absolute value sign is negative. Therefore, g(x) = -x/2 - 3/4.
- When -3/2 < x < 3/2, the expression inside the absolute value sign is positive. Thus, g(x) = x/2 - 3/4.
- When x > 3/2, the expression inside the absolute value sign is once again negative. Therefore, g(x) = -x/2 - 3/4.
3. Plot the points and sketch the graph:
- Plot the key points (3/2, 0) and (-3/2, 0) on the x-axis.
- Sketch a straight line passing through each pair of points: (-∞, -3/4) to (-3/2, 0), (-3/2, 0) to (3/2, -1/2), and (3/2, -1/2) to (∞, -3/4).
The resulting graph should be a V-shaped curve, opening downwards, passing through the points (-3/2, 0) and (3/2, -1/2), and approaching the asymptotes at y = -3/4 as x tends to negative and positive infinity.
Remember to label the x and y axes and any key points or intercepts on the graph.