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

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) (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 . The lowest value possible for t is t=-4 which gives you
.
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).

is pushed one unit TO THE RIGHT from . 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.
-
:
, but g at never touches the y-axis; in any event, find the x-intercept for this branch:

-
:

, which will in the given function intersect the y-axis.
WHERE contact the x-axis:

-
g(x) has only ONE x-intercept, being at x=1.
This is the point (1,0).
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SOLUTION: Confused about graphs here: f(t)=&#8730;(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

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