How do you graph the function #f(x)=-(x+2)^2-5# and its inverse?

To graph this function, we find the #a#,#h#, and #k#.
Since this function is in the form #a(x-h)^2+k#, we can directly find those values.

#a=-1#
#h=-2#
#k=-5#

To graph the points, use the following rule:
For the x values, use the expression #h-2a#,#h-a#,#h#, #h+a#, and #h+2a#. Plug these values in to get the y values.

If we do this, we get:
#(0,-9)(-1,-6)(-2,-5)(-3,-6)(-4,-9)# graph these points to get your answer!

Now, for the inverse.
We first find the inverse of #-(x+2)^2-5#

Remember that when you put the inverse function of the given function into the given function, we get #x#.
So for this case, #-(g(x)+2)^2-5=x# to find the inverse function #g(x)#. (We replace #g(x)# with y for simplicity.)
#-(g(x)+2)^2-5=x#
#-(y+2)^2-5=x#
#-(y+2)^2=x+5#
#(y+2)^2=-x-5#
#y+2=sqrt(-x-5)#
#y=sqrt(-x-5)-2# This is the inverse function.

Notice that any positive value of x cannot work for this function. Only values less than or equal to -5 makes sense.

You could plug in x values less than or equal to -5 into the function to graph it.

Using transformation, you have reflect the graph of #f(x)=sqrtx# to the respect to the y axis.
Then, you have to move it to the left 5 units, then move it down two units.