Consider the differential equation dy/dx=3(x-2)(y-3) where y<0

Expert intro: A horizontal tangent occurs where the derivative is zero. For this differential equation, that happens when either factor in the product is zero. The condition y<0 narrows the possible location to one specific quadrant.

Detailed walkthrough

We are given

$\frac{dy}{dx}=3(x-2)(y-3), \qquad y<0.$

A horizontal tangent means

$\frac{dy}{dx}=0.$

So we need

$3(x-2)(y-3)=0.$

This happens when either:

• $x=2$, or
• $y=3$.

But the solution satisfies $y<0$, so it can never have $y=3$. Therefore the horizontal tangent must occur when

$x=2.$

Since $y<0$, the point lies below the x-axis. On the line $x=2$, the part below the x-axis is in Quadrant IV.

$\boxed{\text{Quadrant IV}}$

💡 Pitfall guide

Do not assume the horizontal tangent must occur where both factors are zero. Only one factor needs to be zero. Another common mistake is ignoring the condition $y<0$, which rules out the line $y=3$ completely.

🔄 Real-world variant

If the condition were $y>3$ instead, then the horizontal tangent could occur on the line $y=3$, which lies above the x-axis. If no restriction on $y$ were given, then horizontal tangents could occur anywhere on the vertical line $x=2$ or the horizontal line $y=3$, depending on the particular solution curve.

🔍 Related terms

horizontal tangent, quadrant, differential equation