Question 658·Hard·Math

The graph has a local minimum at $x=3$, so it must turn from falling to rising at that horizontal position. You need the constant $a$ that puts the valley there, not the valley’s height.

Type $a=0$ to start a slider, then type $p(x)=x^3-ax^2+9ax-27$. Desmos shows the curve rising through $x=3$, so this starting value of $a$ doesn’t put a valley there.

Drag $a$ toward negative values until the valley reaches $x=3$. Then type $-9$ into the slider and click the marked minimum. Desmos labels it $(3,-162)$. The valley’s $x$-coordinate is the condition; its height isn’t the answer. The required constant is $-9$. Grid in -9.

Let $x=3+t$, so $t$ measures how far an input is from $3$. Substitute this into the given polynomial:

Expand the powers:

Group terms by their power of $t$:

At $t=0$, this gives $p(3)=18a$. Subtract that height: $p(3+t)-p(3)=(27+3a)t+(9-a)t^2+t^3$. This difference tells you whether a nearby point is lower than the point at $3$.

For a local minimum, nearby points on both sides of $3$ must be higher. If $27+3a$ weren’t zero, its term $(27+3a)t$ would make the difference negative on one side: $t$ changes sign, while the $t^2$ and $t^3$ terms shrink faster near $0$. So:

Subtract $27$:

Divide by $3$:

With this value, the height difference is $t^2(18+t)$, which is positive for small nonzero $t$. So the point at $x=3$ really is a local minimum when $a=-9$. Grid in -9.