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Section 2.4 #17: Determine whether the graph of the equation 3x2−2y2=3 is symmetric with respect to the x-axis, the y-axis, and/or the origin.
Solution:
test for symmetry with respect to x-axis: Replace y with −y to get the equation
3x2−2(−y)2=3.
Since (−y)2=(−y)(−y)=y2 this equation is equivalent to
3x2−2y2=3.
This equation is the same as the original, so the grpah is symmetric with respect to the x-axis.
test for symmetry with respect to y-axis: Replace x with −x to get the equation
3(−x)2−2y2=3.
Since (−x)2=x2, this equation is equivalent to
3x2−2y2=3,
which is the same as the original. Hence the graph is symmetric with respect to the y-axis.
test for symmetry with respect to origin Replace x with −x and y with −y to get
3(−x)2−2(−y)2=3.
Since (−x)2=x2 and (−y)2=y2 this is equivalent to
3x2−2y2=3,
which is the original equation. Hence the graph is symmetric with respect to the origin.
Section 2.5 #11: Describe how the graph of f(x)=12|x|−2 can be obtained and graph it.
Solution: We start with the graph of y=|x|. We then apply a vertical stretch of 12 (by multiplying y-value by 12). After that we apply a vertical shift down by 2 (i.e. subtract 2 from each y-value).
Section 2.5 #37: The point (−12,4) is on the graph of y=f(x). Find the corresponding point on the graph of y=g(x) where g(x)=12f(x).
Solution: The factor of 12 tells us there is a vertical compression by a factor of 12, meaning that we multiply y-values by 12. Doing so to the point (−12,4) yields the point (−12,2).
Section 2.5 #50: Write an equation for a function that has a graph with the following characteristics: the shape of y=√x, but shifted left 6 units and down 5 units.
Solution: Start with y=√x. Shifting it left by 6 would give us y=√x+6. Shifting that down by 5 units gives us y=√x+6−5.
Section 2.5 #63: The graph of y=f(x) is given as follows:
Now graph y=g(x) where g(x)=−12f(x−1)+3.
Solution: