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secondary 4 | E Maths
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Haven’t done these qns for a while i forgot how already
Date Posted: 2 years ago
Views: 324
the negative 2 not part of qns
Without the "-2", the question won't be solvable. We can only say (without the "-2") that...
y = a (x + 3)² + 5 where a is a negative constant
If we have (-2, 0) as coordinates, you can substitute these values in.
0 = a (-2 + 3)² + 5
0 = a + 5
-5 = a
y = -5 (x + 3)² + 5
y = -5 (x² + 6x + 9) + 5
y = -5x² - 30x - 40
y = a (x + 3)² + 5 where a is a negative constant
If we have (-2, 0) as coordinates, you can substitute these values in.
0 = a (-2 + 3)² + 5
0 = a + 5
-5 = a
y = -5 (x + 3)² + 5
y = -5 (x² + 6x + 9) + 5
y = -5x² - 30x - 40
I guess might be qns error because it says where a b c are constants and also asks for x-intercepts in part 2
thanks for the help
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Date Posted:
2 years ago
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