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

1 Answer
Nov 9, 2017

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]}