J's answer to A's Junior College 2 H2 Maths Singapore question.
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y = ax + bx³
Sub (2,-4), (i.e when x = 2, y = -4 ,)
-4 = a(2) + b(2³)
-4 = 2a + 8b
2a = -4 - 8b
a = -2 - 4b ①
To find gradient of tangent at this point, differentiate y.
dy/dx = a + 3bx²
when gradient of tangent at (2,-4) = 6,dy/dx = 6
So 6 = a + 3b(2²)
6 = a + 12b
a = 6 - 12b
Sub a = 6 - 12b into ①,
6 - 12b = -2 - 4b
6 + 2 = -4b + 12b
8 = 8b
b = 1
Then a = 6 - 12(1) = -6
or
a = -2 - 4(1) = -6
Sub (2,-4), (i.e when x = 2, y = -4 ,)
-4 = a(2) + b(2³)
-4 = 2a + 8b
2a = -4 - 8b
a = -2 - 4b ①
To find gradient of tangent at this point, differentiate y.
dy/dx = a + 3bx²
when gradient of tangent at (2,-4) = 6,dy/dx = 6
So 6 = a + 3b(2²)
6 = a + 12b
a = 6 - 12b
Sub a = 6 - 12b into ①,
6 - 12b = -2 - 4b
6 + 2 = -4b + 12b
8 = 8b
b = 1
Then a = 6 - 12(1) = -6
or
a = -2 - 4(1) = -6
Date Posted:
3 years ago
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