Question

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?

Hint:

Always expand the quadratic and use the vertex formula. For upward parabolas, vertex gives the minimum; for downward parabolas, it gives the maximum.

Answer 1

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:
\[ \boxed{-\tfrac{1}{2}} \]