SOLUTION: Confused about graphs here: f(t)=√(t+4) - 1 I understand how to find its x and y intercepts and its shape. But how do I find where it stops extending on the lefthand side

Algebra ->  Finance -> SOLUTION: Confused about graphs here: f(t)=√(t+4) - 1 I understand how to find its x and y intercepts and its shape. But how do I find where it stops extending on the lefthand side      Log On


   

Question 947843: Confused about graphs here:
f(t)=√(t+4) - 1
I understand how to find its x and y intercepts and its shape. But how do I find where it stops extending on the lefthand side. Meaning the point: (-4,-1) where is that derived from?

Also, g(x)=2|x-1| I know its shape and I know it's pushed to the left +1 and that its y-int is (0,2) but what if I wanted to find the x-int how would that be achieved?
Thanks!
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Not know what you ask for f(t)... Do you mean, what is the domain? The domain is t%3E=-4. The lowest value possible for t is t=-4 which gives you
f%28-4%29=sqrt%28-4%2B4%29-1=-1.
The point for f(t) at the lowest input value of its domain is (-4,-1). This is as far to the left as possible on the graph for f(t).



g%28x%29=2abs%28x-1%29 is pushed one unit TO THE RIGHT from y=2ab%28x%29. g(x) may have two x-intercepts. Use the definition for absolute value to examine x-1.
-
The critical value of x is +1. The sign of x-1 changes there.
-
x%3E1:
g%28x%29=2x-2, but g at x%3E1 never touches the y-axis; in any event, find the x-intercept for this branch: 0=2x-2
2x=2
x=1
-
x%3C1:
g%28x%29=2%281-x%29
g%28x%29=2-2x
g%28x%29=-2x%2B2, which will in the given function intersect the y-axis.
WHERE contact the x-axis:
-2x%2B2=0
2x-2=0
x=1
-
g(x) has only ONE x-intercept, being at x=1.
This is the point (1,0).