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]}

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