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.
Question 1 Report
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.
Answer Details
The gradient function is the derivative. For \(y = ax^2 + bx + c\),
\[\frac{dy}{dx} = 2ax + b.\]
We are told this equals \(8x + 4\), so comparing coefficients:
\[2a = 8 \Rightarrow a = 4, \qquad b = 4.\]
Using the minimum value. At the minimum the gradient is zero:
\[8x + 4 = 0 \Rightarrow x = -\tfrac{1}{2}.\]
The minimum value of \(y\) there is \(1\). Substitute \(a=4,\ b=4,\ x=-\tfrac12\):
\[y = 4\left(-\tfrac12\right)^2 + 4\left(-\tfrac12\right) + c = 4\cdot\tfrac14 - 2 + c = 1 - 2 + c = c - 1.\]
Set \(c - 1 = 1 \Rightarrow c = 2\).
Answer: \(a = 4,\; b = 4,\; c = 2\), giving \(y = 4x^2 + 4x + 2\).
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