How do you graph # y=-2(x-2)^2-4#?

1 Answer
Nov 1, 2017

Graph:
graph{-2(x-2)^2-4 [-6.54, 13.46, -12.2, -2.2]}

See explanation below.

Explanation:

There are more rigorous ways to draw the graph of an parabola by hand (using calculus, mostly), but for our purposes, here's what we're going to do:

Step 1: Identify the Vertex
This is just because you have your parabola in vertex form, which makes this process very easy. For a parabola in vertex form #y = a(x-h) + k#, the vertex is simply #(h, k)#. Therefore, your vertex would be #(2, -4)#.

Step 2: Identify Intercepts
This is #x# intercepts (where #y = 0#), and #y# intercepts (where #x=0#). Let's find these:

#x#- intercepts:

#0 = -2(x-2)^2-4#
#4 = -2(x-2)^2#
#-2 = (x-2)^2#

Now we can stop right there, as we can see that we're gonna end up with the square root of a negative number. Hence, we have no real #x#-intercepts.

#y#-intercepts (these are considerably easier):

#y = -2(0-2)^2-4#
#y = -8#

Step 3: Identify The Direction of the Parabola

This is pretty straightforward -- this basically depends on the sign of the #a# value at the front of your equation. In our case it is negative, so our parabola is going to be pointed downward.

Step 4: Easy Points

This just means you plug in some values of #x# for which it is pretty straightforward to find the value of #y#. #2#, for example, works really well since it cancels out everything under the square, and leaves you with #4#. #1# also works out pretty well, and gives you #-6#.

Now, you just put everything together, and see what turns out:
graph{-2(x-2)^2-4 [-6.54, 13.46, -12.2, -2.2]}

Does that have everything we just calculated?

Hope that helped :)