Let's solve the problem **step by step**: --- ### **Given:** \[ v(T) = T^2 - T - 6 \] Factored as: \[ v(T) = (T-3)(T+2) \] Time period: \[ 1 \leq T \leq 4 \] Intervals to consider: - \([1, 3]\) - \([3, 4]\) Question: **What is happening?** --- ### **Step 1: Understanding \(v(T)\)** - \(v(T)\) represents some variable as a function of time (possibly velocity). - The zeros of \(v(T)\) are important (where it crosses the T-axis). Set \(v(T) = 0\): \[ T^2 - T - 6 = 0 \implies (T-3)(T+2) = 0 \] So, \[ T = 3 \quad \text{or} \quad T = -2 \] But our interval is \(1 \leq T \leq 4\), so **only** \(T = 3\) matters. --- ### **Step 2: Analyze Intervals** #### **a) For \(T \in [1, 3]\):** Pick a value, say \(T = 2\): \[ v(2) = (2-3)(2+2) = (-1)(4) = -4 \] So, \(v(T) < 0\) in this interval. #### **b) For \(T \in [3, 4]\):** Pick a value, say \(T = 4\): \[ v(4) = (4-3)(4+2) = (1)(6) = 6 \] So, \(v(T) > 0\) in this interval. --- ### **Step 3: What is happening at \(T = 3\)?** - At \(T = 3\), \(v(3) = 0\). - This is a **zero crossing point**: the function changes sign. --- ### **Step 4: Interpretation** If \(v(T)\) represents velocity: - **From \(T = 1\) to \(T = 3\), velocity is negative** (moving in the negative direction). - **At \(T = 3\), velocity is zero** (object stops momentarily). - **From \(T = 3\) to \(T = 4\), velocity is positive** (object starts moving in the positive direction). **This means the object changes direction at \(T = 3\).** --- ## **Final Answer:** ### **On the interval \(T \in [1, 3]\), \(v(T)\) is negative, so the object moves in the negative direction.** ### **At \(T = 3\), \(v(T)\) is zero, so the object stops momentarily.** ### **On the interval \(T \in [3, 4]\), \(v(T)\) is positive, so the object moves in the positive direction.** ### **In summary: The object changes direction at \(T = 3\).**