Which transformation can NOT be used to prove that ABC is congruent to DEF
1 answer:
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Answer:
Step-by-step explanation:
In essence, one needs to work their way backwards to solve this problem. Use the information to construct the function.
The function has verticle asymptotes at (x = 4) and (x = 5). This means that the denominator must have (x - 4) and (x - 5) in it. This is because a verticle asymptote indicates that the function cannot have a value at these points, the function jumps at these points. This is because the denominator of a fraction cannot be (0), the values (x - 4) and (x - 5) ensure this. Since if (x) equals (4) or (5) in this situation, the denominator would be (0) because of the zero product property (this states that any number times zero equals zero). So far we have assembled the function as the following:
The function has x-intercepts at (6, 0), and (0, 10). This means that the numerator must equal (0) when (x) is (6) or (10). Using similar logic that was applied to find the denominator, one can conclude that the numerator must be (). Now one has this much of the function assembled
Finally one has to include the y-intercept of (0, 120). Currently, the y-intercept is (60). This is found by multiplying the constants together. (6 * 10) equals (60). One has to multiply this by (2) to get (120). Therefore, one must multiply the numerator by (2) in order to make the y-intercept (120). Thus the final function is the following:
We shall proceed in graphical method
Graph all 4 functions
- Graph 4
Reason:-
It depends upon coefficient
In graph 4 the coefficient outside root is negative so it is decreasing
