I have the question y=-|x^2 -9|

I am asked for the piecewise function. I have written
y={-x^2 +9 if x less than equal to -3}
{ x^2 -9 if x is greater than equal to 3}
I am wondering if I have my are signs backwards.My reasons for choosing the signs I have is if I plug a negative number into the first one and square it, it will be positive and then the negative out front will make it negative. Is there a trick to getting the signs right?

Not only are the signs backwards, but the intervals are incorrect as well.

Take a look at the graph and see whether you can fix it.

http://www.wolframalpha.com/input/?i=-|x^2-9|+for+-5+%3C%3D+x+%3C%3D+5

The trick to getting signs right is remembering that

|x| = x if x >= 0
|x| = -x if x < 0

To determine the signs for the piecewise function, let's break down the expression step by step:

1. First, consider the inner function |x^2 - 9|. This represents the absolute value of the quantity (x^2 - 9).

2. The absolute value function always yields a non-negative value, meaning it is never negative. Therefore, the expression |x^2 - 9| is always greater than or equal to zero.

3. Next, look at the outer function y = -|x^2 - 9|. The negative sign in front of the absolute value function indicates a reflection of the graph over the x-axis. It means that wherever the absolute value function would be positive, the overall function will be negative, and vice versa.

Based on these observations, let's analyze each part of the piecewise function:

For x ≤ -3:
Substituting x = -3 into the expression x^2 - 9, results in (-3)^2 - 9 = 0. So, the value of y in this range is y = -(-3)^2 + 9 = -0 + 9 = 9. Therefore, the first part of the piecewise function should be y = 9 for x ≤ -3.

For x ≥ 3:
Substituting x = 3 into the expression x^2 - 9, yields (3)^2 - 9 = 0. Hence, the value of y in this interval is y = -(3)^2 + 9 = -0 + 9 = 9. Thus, the second part of the piecewise function should be y = 9 for x ≥ 3.

Therefore, the corrected piecewise function would be:
y = 9 if x ≤ -3
y = 9 if x ≥ 3

Notice that the negative sign in front of the absolute value function does not change the sign of either the positive or negative results from the absolute value. It simply reflects the graph over the x-axis.