Answer:
f(x)=(x-1)^2+5 with domain x>1 and range y>5 has inverse g(x)=sqrt(x-5)+1 with domain x>5 and range y>1.
Step-by-step explanation:
The function is a parabola when graphed. It is in vertex form f(x)=a(x-h)^2+k where (h,k) is vertex and a tells us if it's reflected or not or if it's stretched. The thing we need to notice is the vertex because if we cut the graph with a vertical line here the curve will be one to one. So the vertex is (1,5). Let's restrict the domain so x >1.
* if x>1, then x-1>0.
* Also since the parabola opens up, then y>5.
So let's solve y=(x-1)^2+5 for x.
Subtract 5 on both sides:
y-5=(x-1)^2
Take square root of both sides:
Plus/minus sqrt(y-5)=x-1
We want x-1>0:
Sqrt(y-5)=x-1
Add 1 on both sides:
Sqrt(y-5)+1=x
Swap x and y:
Sqrt(x-5)+1=y
x>5
y>1
tenths because 42.02 it would be tenths the first decimal is tenths and second is hundreths
