Question

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

Answer 1

Maximum turning point at (3,1) and y-intercept at (0,-8)

Explanation:

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]}