Given the function \(y(x)=(x+3)(x-2)\), for \(-4<x<4\). What is the value of \(x\) at which the function has a minimum?
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Solution and Explanation
Step 1: Expand the quadratic.
\[
y(x) = (x+3)(x-2) = x^2 + 3x - 2x - 6 = x^2 + x - 6.
\]
Step 2: Nature of parabola.
Coefficient of \(x^2\) is positive (\(+1\)), so it is an upward-opening parabola.
Hence, it has a single global minimum at its vertex.
Step 3: Find vertex.
For quadratic \(y=ax^2+bx+c\), vertex \(x=-\tfrac{b}{2a}\).
Here \(a=1, b=1\):
\[
x=-\frac{1}{2(1)}=-\frac{1}{2}.
\]
Step 4: Check domain.
Domain is \(-4<x<4\). Since \(-0.5\) lies inside, that is the minimum point.
Final Answer:
\[
\boxed{-\tfrac{1}{2}}
\]
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