A construction worker tosses a brick from a tall building. The brick's height (in meters above the ground) \( t \) seconds after being thrown is modeled by \[ h(t)=-5 t^{2}+20 t+105 \] Suppose we want to know the height of the brick above the ground at its highest point. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. \[ h(t)= \] \( \square \) 2) At its highest point, how far above the ground was the brick? \( \square \) meters

We start with the given function:   h(t) = –5t² + 20t + 105 Step 1: Rewrite the function in vertex form. To do this, we complete the square. 1. Factor out the coefficient of t² from the t terms:   h(t) = –5(t² – 4t) + 105 2. Complete the square for the expression in parentheses:   Take half of –4 (which is –2), square it (getting 4), and add and subtract 4 inside the parentheses.   h(t) = –5[(t² – 4t + 4) – 4] + 105 3. Rewrite the quadratic as a perfect square:   h(t) = –5[(t – 2)² – 4] + 105 4. Distribute –5:   h(t) = –5(t – 2)² + 20 + 105 5. Combine the constants:   h(t) = –5(t – 2)² + 125 Thus, the rewritten form is:   h(t) = –5(t – 2)² + 125 Step 2: Determine the maximum height. In the vertex form, the vertex is (2, 125), where t = 2 seconds and the maximum height is 125 meters. Answers: 1) h(t) = –5(t – 2)² + 125 2) The brick reaches a maximum height of 125 meters above the ground.