Question 1165235
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This is the graph of the piecewise function.
{{{
drawing(400,400,-15,15,-15,15,
graph(400,400,-15,15,-15,15,((sqrt(15.80063-x^2))/(sqrt(15.80063-x^2)))*(-x^2+10)),
red(
circle(-4,-6,0.29),
circle(-4,-6,0.31),
circle(-4,-6,0.33),
circle(-4,-6,0.35),
circle(-4,-6,0.37),
circle(-4,-6,0.39),
circle(-4,-6,0.41),
circle(-4,-6,0.43),
circle(-4,-6,0.45),
circle(-4,-6,0.47),
circle(-4,-6,0.49),

circle(4,-6,0.45),
circle(4,-6,0.47),
circle(4,-6,0.49)
),

blue(
line(4,-9,20,-25),
circle(4,-9,0.25),
circle(4,-9,0.27),
circle(4,-9,0.29),
circle(4,-9,0.31),
circle(4,-9,0.33),
circle(4,-9,0.35),
circle(4,-9,0.37),
circle(4,-9,0.39),
circle(4,-9,0.41),
circle(4,-9,0.43),
circle(4,-9,0.45),
circle(4,-9,0.47),
circle(4,-9,0.49)
),

red(locate(-7,-7,"(-4,-6)")),
red(locate(4.5,-5,"(4,-6)")),
blue(locate(4.5,-8,"(4,-9)"))
)
}}}
The red parabola is due to the piece y = -x^2+10, when graphed on the interval -4 <= x < 4.
Graph y = -x^2+10 as normal, then erase the portions when x < -4 or x > 4.
The vertex is at (0,10). 
Two other points on this parabola are (-1,9) and (1,9).


Two graphing tools I recommend are <a href="https://www.desmos.com/calculator">Desmos</a> and <a href="https://www.geogebra.org/graphing?lang=en">GeoGebra</a>
Or you can use something like a TI83.
Another alternative is to use graph paper to do everything by hand.


The blue line is the piece y = -5-x on the interval x >= 4.
Two points on this blue line are (4,-9) and (5,-10).


There is a red closed endpoint at (-4,-6)
There is an open hole at (4,-6) marked in red.
There is a blue closed filled in endpoint at (4,-9)
Another way of saying "closed endpoint" could be "filled in endpoint".


The open endpoint is due to the lack of "or equal to" in the inequality sign. 
This is why we exclude x = 4 from the red parabola.
In contrast every other endpoint is included because there is an "or equal to".


We can clearly see the graph is <u>not</u> continuous. 
There's a jump at x = 4 when going from the red curve to the blue line.


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To determine this algebraically, without needing a graph, plug x = 4 into both pieces to see what results.
I'll label the pieces g(x) and h(x).
g(x) = -x^2+10
g(4) = -4^2+10
g(4) = -16+10
g(4) = -6
This leads to the location (4,-6) which is the red open hole on the graph.


And,
h(x) = -5-x
h(4) = -5-4
h(4) = -9
This leads to the blue filled in endpoint at (4,-9)
The results -6 and -9 do not agree on the same number, so this is a non-graph confirmation that the piecewise function is <u>not</u> continuous.
To be continuous, we would need g(4) = h(4) to be the case. 



Similar practice problems are at the following links
<a href="https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1206484.html">https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1206484.html</a>
<a href="https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1188881.html">https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.1188881.html</a>
<a href="https://www.algebra.com/algebra/homework/Rational-functions/Rational-functions.faq.question.1200349.html">https://www.algebra.com/algebra/homework/Rational-functions/Rational-functions.faq.question.1200349.html</a>
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