![]() Calculating (4,y) and (3,y) and then using symmetry to also plot (6,y) and (7,y) is another way to graph the parabola.Ĭonsider the given domain intervals, for the piecewise graph. Y=a(x-h)^2+k is also a vertex form, for parabolas. The graph of (x-5)^2-2 is the graph of x^2 shifted right five units and down two units. The intercepts are easy to calculate, so plotting the vertex with points (0,0) and (2,0) is another way to graph the absolute-value function. The graph of 2|x-1|-2 is the graph of |x| compressed horizontally by a factor of 2 and shifted right one unit and down two units.Īlso, y=a|x-h|+k is vertex form, and that shows our vertex point (h,k) is at (1,-2). We can see the x-intercept by inspection, then by symmetry we know (-1,2) also. The graph of -x^3+1 is the graph of x^3 reflected across the x-axis and shifted up one unit. For example, for the first equation, should I pick x values less than 1 after I get (1,0) and place the open circle there? Then for those values less than 1, do I substitute it back into the equation? For example, if I pick x value -1, do I substitute it like this: -(-1)^3+1 to get the y-value? Does anyone have a better approach to this? And if possible, can you direct me towards a good resource? So should I make multiple tables instead? (Even though I made a table, my final product didn’t match the key still, so maybe it’s an error on my part). ![]() But since the key did not match this, I had a thought that my process doesn’t work for certain functions. I did the same thing for other equations, substituted the restriction number and got the coordinate. Then I followed the restriction, put an open circle on the point (1,0) and and arrow pointing to everything on the left of (1,0). ![]() For example, for the first equation, I substituted 1 like this: (-1)^3+1 and got the coordinate (1,0). View attachment 28865What I initially thought to do was substitute the restriction’s number into the respective equation, but my final graphed piecewise didn’t match the key. ![]() What I initially thought to do was substitute the restriction’s number into the respective equation, but my final graphed piecewise didn’t match the key. This was on a friend’s precalculus packet:
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