Given : f(x)= 3|x-2| -5
f(x) is translated 3 units down and 4 units to the left
If any function is translated down then we subtract the units at the end
If any function is translated left then we add the units with x inside the absolute sign
f(x)= 3|x-2| -5
f(x) is translated 3 units down
subtract 3 at the end, so f(x) becomes
f(x)= 3|x-2| -5 -3
f(x) is translated 4 units to the left
Add 4 with x inside the absolute sign, f(x) becomes
f(x)= 3|x-2 + 4| -5 -3
We simplify it and replace f(x) by g(x)
g(x) = 3|x + 2| - 8
a= 3, h = -2 , k = -8
Answer:
theres no picture
Step-by-step explanation:
The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at
(-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer:

Step-by-step explanation:
This can be solved with Pythagreon Thereom.

Follow the steps below. A and B are the heaight an base, C is they hypotonuese.
