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

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

Step-by-step explanation:

Consider the provided equation:

The vertex form of a quadratic is:

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.

is there supposed to be a diagram or a graph for this one? ?

step-by-step explanation:

show us the triangle so we can answer it