The function is definately defined at x=0 but not x=1.
But its just one part of the coordinate (x,y).
If the value of y or f(x) is considered, you'll see that it is never possible to attain f(x)=0. In other terms (x,y)= (0,0) is not a defined point in the graph of the function because the graph doesnt pass through that point.
Now I hope you understood what I meant!
Conclusion- The above function is not defined at all points in the space having the abscissa or x=1 in the coordinate and also at ordinate or y=0 in the coordinate.nation:
Answer:
B. y =
Step-by-step explanation:
I. Given x = 15 and y = 4
4 = 
4 = 2+2
4 = 4 true
II. Give x = 0 and y = 2
2 = 
2 = 0 + 2
2 = 2 true
III. Given x = -15 and y = 0
0 = 
0 = -2 + 2
0 = 0 true
Good luck.
General Idea:
W
hen we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)
When
, then P' is enlargement of P
When
, then P' is reduction of P
Applying the concept:
In the diagram given A'B'C' is enlargement of ABC. So the correct option will be option A.
, enlargement
Answer:
A
Step-by-step explanation: