How do you sketch the graph of #y=-2(x-1)^2+7# and describe the transformation?

Question

Isaiah Anthony · Accepted Answer

See method below.

Transformation Let #f(x)=x^2# #f(x-1)=(x-1)^2# This is a translation by 1 to the right. #2f(x-1)=2(x-1)^2# This is a vertical stretch, scale factor 2. #-2f(x-1)=-2(x-1)^2# This is a reflection in the x-axis (y=0). #-2f(x-1)+7=-2(x-1)^2+7# This is a translation up by 7. Put this together; our transformation is: Reflection in the x-axis, vetical stretch scale factor 2, transformation by #((1),(7))#. Sketching We need to find some points to plot the curve. #y=-2(x-1)^2+7# From this completed square form, we can tell that the minimum will be at #(1, 7)# Find the x-intercepts: Let #y=0# #0=-2(x-1)^2+7# #-7=-2(x-1)^2# #7/2=(x-1)^2# #+-sqrt(7/2)=x-1# #x=1+-sqrt(7/2)# so #x=1+sqrt(7/2)# or #x=1-sqrt(7/2)# Find the y-intercepts: #y=-2(x^2-2x+1)+7# #y=-2x^2+4x-2+7# #y=-2x^2+4x+5# Let #x=0 => y=5# So there is a y-intercept at #(0, 5)# The co-efficient of #x^2# is negative, so it is an upside-down-U-shaped curve. The graph should look like this: graph{-2(x-1)^2+7 [-14.16, 17.88, -6.48, 9.54]}

Kennedy Adams · Answer

undefined To sketch the graph of $ y = -2(x-1)^2 + 7 $ and describe the transformation:

1. Start with the basic graph of $ y = x^2 $, which is a parabola opening upwards with its vertex at the origin.

2. Apply the transformation $ (x-1)^2 $, which shifts the graph one unit to the right.

3. Apply the multiplication by -2, which reflects the graph over the x-axis and changes its direction to open downwards.

4. Apply the addition of 7, which shifts the graph vertically upwards by 7 units.

5. The resulting graph is a downward-opening parabola with its vertex at (1, 7).