X-5+5x-12=25
x+5x=25+5+12
6x=42
x=42/6
x=7
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:
a = 4
Step-by-step explanation:
y = 2x + 1 ......(1)
Going through y - y1 = m(x - x1)
m is slope and it is 2
y1 = 9 and x1 = a
y - 9 = 2(x - a)
y - 9 = 2x - 2a
y = 2x - 2a + 9 ........(2)
Equating (1) and (2)
2x + 1 = 2x - 2a + 9
Collecting like terms
2x - 2x + 2a = 9 - 1
2a = 8
a = 8/2
a = 4
For this case, the parent function is given by:

We apply the following function transformation:
Vertical compressions:
To graph y = a * f (x)
If 0 <a <1, the graph of y = f (x) is compressed vertically by a factor a. (Shrinks)
We have then:

Horizontal translations:
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:

Vertical translations:
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
Answer:
Vertical compression by factor of 1/2. Horizontal displacement 3 units to the left. Vertical displacement 2 units up.
Answer:
2.26666666667 is the answer