Question

analyzing graphs
which interval for the graphed function has a local minimum of 0?
options: $-3, -2$, $2, 4$, $-2, 0$, $1, 2$
graph with points $(-2.5, 0)$, $(-1.5, -6)$, $(0, 0)$, $(1.2, 2.9)$, $(3, 0)$

Explanation:

Step1: Recall local minimum definition

A local minimum is a point where the function value is less than or equal to neighboring points. We need to check which interval contains a point with \( y = 0 \) as a local minimum.

Step2: Analyze each interval

- For \([-3, -2]\): The point \((-2.5, 0)\) is on the graph here, but the function is decreasing then increasing? Wait, no—wait, the graph at \( x=-2.5 \) is a root, but the local minimum at \((-1.5, -6)\) is below. Wait, no, we need a local minimum of 0. Wait, the points with \( y=0 \) are \((-2.5, 0)\), \((0, 0)\), \((3, 0)\). Now, check intervals:
- \([-3, -2]\): Contains \((-2.5, 0)\). Is this a local minimum? The graph around \( x=-2.5 \): left of \( x=-2.5 \), the function is decreasing (going down from left), then at \( x=-2.5 \), then after, does it go up? Wait, the local minimum is at \((-1.5, -6)\), so \((-2.5, 0)\) is a root, but is it a local min? Wait, no—wait, the interval \([2, 4]\): contains \((3, 0)\). At \( x=3 \), the function has a minimum (since after \( x=3 \), it goes up). Wait, no, let's check the graph: the point \((3, 0)\) is a local minimum? Wait, the graph at \( x=3 \): before \( x=3 \), the function is decreasing (from \( (1.2, 2.9) \) down to \( (3, 0) \)), then after \( x=3 \), it goes up. So \( (3, 0) \) is a local minimum. And \( 3 \) is in \([2, 4]\). Wait, but wait, the other points: \((0, 0)\) is a root, but around \( x=0 \), the function comes from below (at \( (-1.5, -6) \)) up through \( (0, 0) \), so \( (0, 0) \) is a local maximum? No, wait, the graph: from \( (-1.5, -6) \) (local min) up to \( (1.2, 2.9) \) (local max), then down to \( (3, 0) \) (local min), then up. So the local minima with \( y=0 \) is at \( x=3 \), which is in \([2, 4]\). Wait, but wait, the interval \([-3, -2]\): the point \((-2.5, 0)\) – before \( x=-2.5 \), the function is decreasing (from left, going down), then at \( x=-2.5 \), then after, does it go up? Wait, the local minimum is at \( (-1.5, -6) \), so \( (-2.5, 0) \) is a root, but the function at \( x < -2.5 \) is decreasing (going down), then at \( x=-2.5 \), then after \( x=-2.5 \), it goes down to \( (-1.5, -6) \)? Wait, no, the graph: leftmost part, the function is coming from up, then goes down to \( (-1.5, -6) \), then up to \( (0, 0) \), etc. Wait, maybe I misread. Let's re-express:

The graph has a local minimum at \((-1.5, -6)\) (y=-6), then a local maximum at \((1.2, 2.9)\) (y=2.9), then a local minimum at \((3, 0)\) (y=0), and roots at \((-2.5, 0)\), \((0, 0)\), \((3, 0)\).

So we need an interval where the local minimum is 0. The local minimum at \((3, 0)\) is in the interval \([2, 4]\). Let's check the options:

- \([-3, -2]\): contains \((-2.5, 0)\), but the function at \( x=-2.5 \) is a root, but the local minimum near there is \((-1.5, -6)\), so \( y=-6 \), not 0.
- \([2, 4]\): contains \( x=3 \), where the local minimum is \( y=0 \).
- \([-2, 0]\): contains \((0, 0)\), but around \( x=0 \), the function is increasing from \((-1.5, -6)\) to \((1.2, 2.9)\), so \( (0, 0) \) is not a local minimum (it's a root, but the function is increasing through it, so local max? No, local min would be where it's the lowest around. At \( x=0 \), the function is going up, so it's a local minimum? Wait, no: from \( x=-1.5 \) (y=-6) to \( x=1.2 \) (y=2.9), the function is increasing, so \( x=0 \) is on the increasing part, so the local minimum in \([-2, 0]\) is \((-1.5, -6)\), not 0.
- \([1, 2]\): the function is decreasing from \((1.2, 2.9)\) to \((3, 0)\), so in \([1, 2]\), the function is decreasing, so the minimum in \([1, 2]\) is at \( x=2 \), but \( y \) there is not 0 (since at \( x=3 \) it's 0, so at \( x=2 \), \( y \) is between 2.9 and 0, not 0).

So the interval with a local minimum of 0 is \([2, 4]\).

Answer:

\([2, 4]\)