Best answer:
Your update is rather confusing, I think you meant
y = x³ - 3x² + 2x + 4
y = x³ - 3x² + 2x + 4
y' = 3x² - 6x + 2
If the tangents are parallel, then y' is equal at both locations, i.e.
3a² - 6a + 2 = 3b² - 6b + 2 = m
Where m is the gradient of the tangent.
3a² - 6a + 2 = m
3a² - 6a + 2 - m = 0
This will...
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Best answer: Your update is rather confusing, I think you meant
y = x³ - 3x² + 2x + 4
y = x³ - 3x² + 2x + 4
y' = 3x² - 6x + 2
If the tangents are parallel, then y' is equal at both locations, i.e.
3a² - 6a + 2 = 3b² - 6b + 2 = m
Where m is the gradient of the tangent.
3a² - 6a + 2 = m
3a² - 6a + 2 - m = 0
This will have 0, 1 or 2 roots. Obviously, 3b² - 6b + 2 = m has the same roots. So if a and b are different, then they must represent the 2 different roots in the 2 roots case.
So we consider the case that 3a² - 6a + 2 - m = 0 has 2 roots.
let those roots be r and s
The sum a + b is then r + s
We can factor 3a² - 6a + 2 - m = 3(a - r)(a - s) = 3a² - 3(r + s)a + 3rs
The coefficient of the a is -3(r + s), so therefore -3(r + s) = -6, or r + s = 2
Therefore a + b = 2
4 answers
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23 hours ago