How do you graph #y = –2(x + 3)^2 + 1#?

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Answer 1 · 2017-12-26T06:47:49

The graph should look like this:

graph{y=-2(x+3)^2+1 [-10, 10, -5, 5]}

First, make a table for #x# and #y#.

Second, plug in values for #x# like #-5,-4,-3,...,0,1,2,3...#

Third, figure out the #y# values from the equation corresponding to the #x# values (in this case it's #y=-2(x+3)^2+1#)

Create an #x,y# graph and plot the points you have found.

Finally, connect the dots to receive a graph like this:

graph{y=-2(x+3)^2+1 [-10, 10, -5, 5]}