\begin{align*}y&\leq (x-3)^2\\y&\geq -x+9\end{align*}

In the xy-plane, a point with coordinates (a, b) lies in the solution set of the system of inequalities above. If a and b are greater than zero, what is the least possible value of a ?



Comments (3)

But, but, but … Can you explain the solution for those who are allergic to graphing? Why is solving this by setting the two inequalities equal to each other (and ending up with solutions x = 2, -3) going to give the wrong answer?

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