Question

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

Answer 1

The two graphs should look like this:
graph{-(x+2)^2-5 [-18.02, 18.03, -9.01, 9.01]}
graph{sqrt(-x-5)-2 [-10, 10, -5, 5]}

Explanation:

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.