Question

21 y=a(x-2)2 +b y =5 in the system of equations above, for which of the following values of a and b does the system have no solution? A) a= 1 and b=-4 B) a 2 and b=5 C) a=-1 and b 6 Da-2 and b 4

Answer 1

The correct option is D. a = -2 and b = 4.

Explanation:

Consider the provided equation:

`y=a(x-2)^2+b\ \text{and}\ y=5`

The vertex form of a quadratic is:

`y= a(x-h)^2+k`

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.