(4p +3)(5p2 + 6) is your answer
I’m not really sure but I would say c
The range is the ouutput from inputing the input
basically
25=k²+2k+1 and 64=k²+2k+1
the values that satisfy both equations (not at the same tim) are the valuess that are the domain
solve each
25=k²+2k+1
minus 25 both sides (or recognize the perfect square trinomial, but anyway)
0=k²+2k-24
factor
0=(k+6)(k-4)
set to zero
k+6=0
k=-6
k-4=0
k=4
k=-6 or 4
64=k²+2k+1
minus 64 both sides
0=k²+2k-63
facor
0=(k-7)(k+9)
set to zer
k-7=0
k=7
k+9=0
k=-9
k=-9 or 7
so the domain has the numbers
-9,-6,4,7
it seems we only want the positive square roots so
answer is {4,7} is the domain
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
