The gradient function of \(y = ax^{2} + bx + c\) is \(8x + 4\). If the function has a minimum value of 1, find the values of a, b and c.
\(y = ax^{2} + bx + c\)
Gradient = \(\frac{\mathrm d y}{\mathrm d x} = 2ax + b = 8x + 4\)
Equating, we have
\(2a = 8 \implies a = 4\)
\(b = 4\)
For minimum value, gradient = 0
\(8x + 4 = 0 \implies x = -\frac{1}{2}\)
At \(x = -\frac{1}{2}, y = 1\)
\(1 = 4(-\frac{1}{2})^{2} + 4(-\frac{1}{2}) + c\)
\(1 = 1 - 2 + c\)
\(1 = c - 1 \implies c = 2\)
\(a, b, c = 4, 4, 2\).
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