Graph $f(x) = (x+2)^2(x-5)$ and look for the lowest point (the valley) on the curve. The x-coordinate of that point is what you need.

This is a cubic function. The factor $(x+2)^2$ means the graph touches (but doesn't cross) the x-axis at $x = -2$. The factor $(x-5)$ means it crosses at $x = 5$. Where might the valley be?

In Desmos, type f(x) = (x+2)^2(x-5) and make sure the viewing window shows roughly from $x = -6$ to $x = 6$.

Look at the graph: you should see the curve touch the x-axis at $x = -2$ (a local high point) and then dip down to a low point before rising again. Zoom in around that valley.

Click on the lowest point of the curve (the valley). Desmos will show its coordinates. The x-coordinate is approximately 2.67. Compare this to the answer intervals and select the one that contains this value.

The function $f(x) = (x+2)^2(x-5)$ has:

• A double zero at $x = -2$ (the graph touches but doesn't cross the x-axis here)
• A simple zero at $x = 5$ (the graph crosses the x-axis here)

Since the leading coefficient is positive (expanding gives $x^3 + .$..), the graph goes down to $-\infty$ on the left and up to $+\infty$ on the right.

Because the graph has a double zero at $x = -2$, it "bounces" off the x-axis there. This creates a local maximum at $x = -2$.

The graph then dips down (going negative) before rising to cross the x-axis at $x = 5$. The lowest point of this dip is the local minimum.

Graph $f(x) = (x+2)^2(x-5)$ in Desmos. You'll see:

• The graph touches the x-axis at $x = -2$ (local maximum)
• The graph dips down to a valley somewhere between $x = -2$ and $x = 5$
• The graph crosses the x-axis at $x = 5$

Click on the lowest point of the valley. Desmos shows the x-coordinate is approximately $x \approx 2.67$.

The x-coordinate of the local minimum is approximately $2.67$, which is $\frac{8}{3}$.

This value lies in the interval $2 < x < 3$, so the answer is C.