How do you graph #y=-2(x-1)^2+1#?

Question

Ethan Adkins · Accepted Answer

With a vertex of #(1,1)#, use #x#-values of #-1,0,2,3# in the equation to determine reference points, creating the graph: graph{-2(x-1)^2+1 [-10, 10, -5, 5]}

Knowing the formula is #f(x)=a(x-h)^2+k#, you can determine that the vertex of the graph will be at point #(1,1)#, as these values translate the entire graph itself. The #a#-value being negative indicates that the vertex is a maximum, making the graph open downward with a vertical stretch factor of #2#. Depending on your teacher, you may be required to include more or fewer reference points in a graph, but the simplest rule is to plugin 2 #x#-values less than your initial point (being the vertex), and 2 #x#-values greater than your initial point to the function. Following this concept, you should get additional points: #(-1,-7), (0,-1), (2,-1),# and#(3,-7)#. It is also helpful to realize that the points are a mirror of one another in their #y#-values, so long as the #x#-values are an equal distance from the #x#-value of the vertex.

Genesis Allen · Answer

undefined To graph the function y = -2(x - 1)^2 + 1, you can follow these steps:
1. Identify the vertex, which is (h, k) in the form (x - h)^2 + k. In this case, the vertex is (1, 1).
2. Since the coefficient of (x - h)^2 determines the direction of the parabola, which is negative in this case, the parabola opens downwards.
3. Plot the vertex (1, 1) on the coordinate plane.
4. Use the symmetry of the parabola to plot other points. Since the parabola is symmetric about its vertex, you can choose points equidistant from the vertex on either side. For example, if you move one unit to the right from the vertex, the corresponding y-value decreases by 2 units (due to the coefficient -2). Similarly, if you move one unit to the left from the vertex, the corresponding y-value decreases by 2 units.
5. Connect the plotted points smoothly to draw the graph of the function.