So you want to graph #f(x)=-7(x-2)^2-9# using only the "important values." The first value we can find is the #vertex# of the graph, or the location where the instantaneous slope is 0. Essentially the highest or lowest point of the graph. Thankfully, your function is vertex form. This means it's represented in the form
#f(x)=a(x-h)^2+k#
Where the vertex is at #(h,k)#. So we can get the first point right off the bat. Your vertex is at #(2,-9)#. Keep note of how I wrote that function. By default, it's #(x-h)#, meaning the sign between them is the opposite sign of the #x# value in the vertex (i.e. Your function says #(x-2)#, which means the #x# value will be 2).
Now, we just need two more points. If the function actually had zeroes, I would use those. But we don't have zeroes, as evidenced by the graph above, as well as the fact the graph opens downwards (#a# is negative) and the vertex is below the #x# axis (#k# is negative). Therefore, we need two other values. The next best value to use would be the #y#-intercept. This can be found by plugging in 0 for #x#.
#-7(x-2)^2-9#
#=-7(0-2)^2-9#
#=-7(-2)^2-9#
#=-7(4)-9#
#=-28-9#
#=-37#
While we know the vertex's position is at #(2,-9)# and the #y#-intercept is at #(0,-37)#, we still need one more point. I'm going for the point on the opposite side of the graph compared to the #y#-intercept. Essentially, the #x# distance from the #y#-intercept to the vertex is 2. Therefore, adding two to the #x# of the vertex will give us the position of our other point at (4,-37). Finally, we can connect these lines to give us the final graph.
graph{-7(x-2)^2-9 [-3, 5, -75, 1]}
Though generally, we want to have the greatest number of important values we can: The roots (zeros), the vertex, and the #y#-intercept.
