Answer:
The correct option is D. a = -2 and b = 4.
Step-by-step explanation:
Consider the provided equation:
[tex]y=a(x-2)^2+b\ \text{and}\ y=5[/tex]
The vertex form of a quadratic is:
[tex]y= a(x-h)^2+k[/tex]
Where, (h,k) is the vertex and the quadratic opens up if 'a' is positive and opens down if 'a' is negative.
Now consider the provided option A. a = 1 and b = -4.
Since the value of a is positive the graph opens up and having vertex (2,-4). Thus graph will intersect the line y = 5.
Refer the figure 1:
Now consider the option B. a = 2 and b = 5.
Since the value of a is positive the graph opens up and having vertex (2,5). Thus graph will intersect the line y = 5.
Refer the figure 2:
Now consider the option C. a = -1 and b = 6.
Since the value of a is negative the graph opens down and having vertex (2,6). Thus graph will intersect the line y = 5.
Refer the figure 3:
Now consider the option D. a = -2 and b = 4.
Since the value of a is negative the graph opens down and having vertex (2,4). Thus graph will not intersect the line y = 5.
Refer the figure 4:
Hence, the correct option is D. a = -2 and b = 4.
