Ben uses a compass and a straightedige to bisect angle PQR, as shown below.
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For a function to be continuous at an x-value of -3 you need to make sure two things line up:
First, we need to show that the limit from the left equals the limit from the right.
Second, we need to show that this limit equals the functions value.
The left hand limit involves the first piece, f(x) = x^2 - 9:
The right hand limit invovles the second piece, f(x) = 0:
Since the two one-sided limits do match, we can just say:
(no one-sided pieces needed now)
So that was step #1, to make sure the limit exists. Next we need to make sure the limit is headed to the same place where the functions. Since we're using x=-3, we'll use the top piece of the function because x=-3 fits with that piece ( x ≤ -3 ).
From this, we know that , so the function is continuous at -3. (We also know parabolas and lines are continuous in general, so we only needed to check where the two pieces came together at x = -3.)
Part (a)
The locus is the set of points that satisfy the given conditions. Namely that the point P needs to be 1 unit from the line given.
Think of a median along a highway. Then imagine the two shoulders of the road. The locus describes the two parallel lines that are to this given median. In other words, the median is y = x+2 and the road shoulders are unknown for now.
Though we do know that parallel lines have equal slopes (but different y intercepts).
Therefore, the answers to the next section will be of the form y = x+k for some real number k.
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Part (b)
Use your favorite graphing tool to plot y = x+2. I recommend GeoGebra since I use it all the time, and it's what I used to make the drawing below. Desmos is another tool you can use. Both are free.
Note that point A(0,2) is on the line y = x+2. The line perpendicular to y = x+2 and goes through (0,2) is y = -x+2. From here, plot a circle centered at (0,2) with radius 1. The equation of this circle is x^2 + (y-2)^2 = 1
The intersection points of the circle and y = -x+2 will provide starting points, so to speak, for the two shoulder lines I mentioned in part (a).
The approximate locations of those points are:
B = (0.71, 1.29)
C = (-0.71, 2.71)
From here, it shouldn't be too hard to find the y = mx+b forms of the locus lines. Let me know if you need help with that, or if you need me to check your answers.
Side note: Why is it called a locus? I'm not sure, but my gut feeling tells me it has to do with the insects of a similar sounding name (locus the math term and locust the insect). Imagine millions of them forming various shapes in the sky, similar to how many points are used in the 2D plane to create various curves.
