MHB Find x- and y- Intercepts....2

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The piecewise-defined function has two segments: y = -x^2 for -2 < x ≤ 0 and y = (x/2) for 0 < x ≤ 4. The x- and y-intercepts for the upper piece occur at the origin (0, 0), as both values equal zero when x is 0. However, for the lower piece, the function is not defined at x = 0, meaning the intercepts do not apply here. Thus, the overall function only has intercepts at (0, 0) from the upper piece. Understanding the defined intervals is crucial for accurately determining intercepts in piecewise functions.
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Find the x- and y-intercepts of the piecewise-defined function.

y = -x^2 if -2 < x < or = 0...this is the upper piece

y = (x/2) if 0 < x < or = 4...this is the bottom piece

For the upper piece:

y = -x^2

Let x = 0

When x = 0, y = 0.

The y-intercept is 0 and it takes place at the origin.

Let y = 0

0 = -x^2

0/-1 = x^2

0 = x^2

sqrt{0} = sqrt{x^2}

0 = x

The x-intercept is 0 and it also takes place at the origin.

For the bottom piece:

y = (x/2)

Let x = 0

When x = 0, y = 0.

The y-intercept is 0 and it takes place at the origin.

Let y = 0

0 = (x/2)

0(2) = (x/2)*2

0 = x

The x-intercept is 0 and it takes place at the origin.

Can I say that the x- and y-intercept for this entire function takes place at the point (0, 0)?
 
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For the bottom piece:

y = (x/2)

Let x = 0

no ... this piece of the function is not defined at x = 0 ... the defined interval is $0 < x \le 4$the x and y intercept at (0,0) is only defined for the quadratic piece, defined over the interval $-2 < x \le 0$
 
I need to work more on piecewise-defined functions.
 
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