How do you graph the function and its inverse of #f(x)=-(x-3)^2+1#?

Let #y=-(x-3)^2+1#

The curve has a minimum point at #(3, 1)# since this is the completed square form.

#y=-(x^2-6x+9)+1#
#y=-x^2+6x-8#

The curve is a 'n' shaped quadratic since the coefficient of #x^2# is negative, so the turning point is a maximum.

let #x=0, y=-8#

From this information, we can put a turning point at #(3,1)# and a y-intercept at #y=-8#. graph{-(x-3)^2+1 [-11.25, 11.25, -5.63, 5.62]}

now for #y=f^-1(x)#

Looking at f(x), for our input x, we go:

#x-> -3 -> Ans^2 -> xx-1 -> +1 -> f(x)#
So
#f(x) -> -1 -> -:-1 -> sqrtAns -> +3 -> f^-1(x)#

So #f^-1(x)=sqrt(-(x-1))+3#

Let #y=sqrt(-(1-x))+3 #

Immediately we get that part of the graph will be imaginary. This part of the graph is where the bit under the square root sign #(-(x-1))<=0#

So let #-(x-1)>=0#
#1-x>=0#
#1>=x#
#x<=1#

So any x values above 1 aren't plotted.

Let #y=0#
#0=sqrt(1-x)+3#
#-3=sqrt(1-x)#
#9=1-x#
#x=-8#
But subbing this into #f^-1(x)# gives us #sqrt(1--8)+3=sqrt9+3=6!=0#

So we have no x-intercepts.

The graph starts when x=1. So let #x=1#

#y=sqrt(1-1)+3#
#y=3#
So we start the graph at (1,3).

To find the y-intercept, let #x=0#
#y=sqrt1+3=4#

Now we draw our graph, starting at (1,3) and passing through (4,0)

graph{sqrt(-(x-1))+3 [-13.875, 8.625, -0.995, 10.255]}