Find x- and y- Intercepts....2

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SUMMARY

The x- and y-intercepts of the piecewise-defined function are both located at the origin (0, 0). The upper piece of the function, defined as y = -x² for the interval -2 < x ≤ 0, confirms that both intercepts occur at this point. The lower piece, y = (x/2) for 0 < x ≤ 4, does not contribute additional intercepts since it is not defined at x = 0. Therefore, the overall intercepts for the entire function are solely at (0, 0).

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mathdad
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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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