How do you sketch the graph of #y=0.5(x-2)^2-2# and describe the transformation?

Question

Abigail Allen · Accepted Answer

The y-intercept is at the origin, #(0, 0)#.
The x-intercepts are
#(0,0)# and #(4,0)#.
The turning point is #(2, -2)#.
graph{ y = 0.5(x-2)^2-2 [-10, 10, -5, 5]}
The graph has shifted 2 units to the right and 2 units down compared to #y=x^2#. The graph is also wider than #y=x^2# since the value of #a# in the equation is #1/2#.
Since the equation is in the form of #y = a(x - h)^2 + k#, the graph's turning point shifts 2 units to the right, and 2 units down. Therefore, the turning point is #(2, -2)#. The next step is to find the intercepts. Recall that: Let #x = 0#: #y = 0.5((0)-2)^2 - 2# #y = 0# The y-intercept is at the origin, #(0, 0)#. Let #y = 0#: #0 = 0.5(x-2)^2 - 2# #0 = 0.5((x-2)^2 - 4)# #0 = (x-2)^2 - 4# #0 = ((x-2) + 2)((x-2) - 2)# #0 = x(x-4)# #x - 4 = 0#, #x = 0# #x = 4#, #x = 0# Therefore, x-intercepts are #(0,0)# and #(4,0)#. Now that you have all the information, let's graph it along with y=x^2. graph{y=0.5(x-2)^2 - 2 [-10, 10, -5, 5]} As you can see, the graph has shifted 2 units to the right and 2 units down compared to #y=x^2#. The graph is also wider than #y=x^2# since the value of #a# in the equation is #1/2#.

Avery Alexander · Answer

undefined To sketch the graph of $ y = 0.5(x-2)^2 - 2 $, follow these steps:

1. Identify the parent function: $ y = x^2 $
2. Determine the transformations:
   - Horizontal shift of 2 units to the right (since $ x-2 $)
   - Vertical compression by a factor of 0.5 (since $ 0.5(x-2)^2 $)
   - Vertical translation of 2 units downwards (since $ -2 $)
3. Plot key points:
   - Vertex: (2, -2)
   - Symmetry: The parabola is symmetric about the vertical line passing through the vertex.
4. Draw the graph: 
   - The graph opens upwards due to the positive coefficient of $ x^2 $.
   - Sketch the parabolic curve through the vertex and other key points.
5. Label the graph and axis if necessary.