Which of the following sequences of transformation is used to obtain figure A’B’C’D’ from ABCD?
1 answer:
Answer:
Most likely (B)
Step-by-step explanation:
Points of ABCD:
A (3,1)
B (3,4)
C (5,5)
D (5,2)
The algebraic rule for reflecting across the y axis:
(x,y) ---> (-x, y)
Points of ABCD after being reflected: (shown by figure 2)
A (-3, 1)
B (-3, 4)
C (-5, 5)
D (-5, 2)
Then, the figure got translated two units to the left, resulting in figure F in the picture, and A’B’C’D’ in the question.
Points of ABCD after being translated by (x-2, y) : (shown by figure F)
A (-5, 1)
B (-5, 4)
C (-7, 5)
D (-7, 2)
This should be the coordinates of A’B’C’D’.
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The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
Armed with the above rules, we can split up the summation into simple terms:
=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14
(c) the maximum value of f(x) is f(14)=196
Armed with the above rules, we can split up the summation into simple terms:
=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14
(c) the maximum value of f(x) is f(14)=196