The function \(f\) is defined by \(f : x \mapsto 6x - x^2 - 5\) for \(x \in \mathbb{R}\).
Given that the line \(y = mx + c\) is a tangent to the curve \(y = f(x)\), show that \(4c = m^2 - 12m + 16\).
Solution
To show that \(4c = m^2 - 12m + 16\), we start by equating the line and the curve: \(mx + c = 6x - x^2 - 5\).
Rearrange to form a quadratic equation: \(x^2 - (6-m)x + (c+5) = 0\).
For the line to be tangent to the curve, the discriminant of this quadratic must be zero: \(b^2 - 4ac = 0\).
Here, \(a = 1\), \(b = -(6-m)\), and \(c = c+5\).
Calculate the discriminant: \((6-m)^2 - 4(1)(c+5) = 0\).
Expand and simplify: \(36 - 12m + m^2 - 4c - 20 = 0\).
Rearrange to find \(4c = m^2 - 12m + 16\).
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