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
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
slope, m = (6-0) /(0--2) = 6/2 = 3
intercept, c at y = -2 when x = 0
Equation
y = mx + c
y = 3x -2
We know that
if PQ <span>is the perpendicular bisector of line MN
then
MQ=QM
and
NP=MP
so
6x+1=10x-17----------> 10x-6x=17+1-----> 4x=18---------> x=18/4----> x=4.5
QN=2x+5-----> QN=2*4.5+5-----> QN=14
the answer is
QN is 14</span>
