Answer:
The correct option is D. <em>a</em> = -2 and <em>b</em> = 4.
Step-by-step explanation:
Consider the provided equation:
![y=a(x-2)^2+b\ \text{and}\ y=5](https://tex.z-dn.net/?f=y%3Da%28x-2%29%5E2%2Bb%5C%20%5Ctext%7Band%7D%5C%20y%3D5)
The vertex form of a quadratic is:
Where, (<em>h</em>,<em>k</em>) is the vertex and the quadratic opens up if '<em>a</em>' is positive and opens down if '<em>a</em>' is negative.
Now consider the provided option A. <em>a</em> = 1 and <em>b</em> = -4.
Since the value of <em>a</em> is positive the graph opens up and having vertex (2,-4). Thus graph will intersect the line <em><u>y</u></em> = 5.
Refer the figure 1:
Now consider the option B. <em>a</em> = 2 and <em>b</em> = 5.
Since the value of <em>a</em> is positive the graph opens up and having vertex (2,5). Thus graph will intersect the line <em>y</em> = 5.
Refer the figure 2:
Now consider the option C. <em>a</em> = -1 and <em>b</em> = 6.
Since the value of <em>a</em> is negative the graph opens down and having vertex (2,6). Thus graph will intersect the line <em>y</em> = 5.
Refer the figure 3:
Now consider the option D. <em>a</em> = -2 and <em>b</em> = 4.
Since the value of <em>a</em> is negative the graph opens down and having vertex (2,4). Thus graph will not intersect the line <em>y</em> = 5.
Refer the figure 4:
Hence, the correct option is D. <em>a</em> = -2 and <em>b</em> = 4.