Question

1. identify the vertex of the parabola represented by the equation: y = 2x² - 8x + 5 a. (3, - 2) b. (4, 1) c. (0, 5) d. (2, - 3) 2. a ball is thrown into the air from the top of a building. the height of the ball can be modeled by the equation h=-16t² + 64t + 80, where h is the height in feet and t is the time in seconds. what is the maximum height the ball reaches? 3. patrick mahomes throws a football with an initial velocity of 27.7 m/s. the ball height above the ground at time (t) seconds is given by the equation h(t)=-4.9t² + 27.7t. how long does it take for the ball to reach the ground? 4. a graph represents the height of a baseball, h, in feet as a function of time, t, in seconds after it was hit by a baseball player with function: h(t)=-16t² + 50t + 4. answer the following: a. what is the initial height of the baseball? b. approximately how many seconds does it take for the ball to reach the maximum height? c. determine the maximum height of the baseball. d. how many seconds does it take for the ball to reach the ground? 5. an object is launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. its height is represented by the function: s(t)=-16t² + 64t + 80 a. what will the objects maximum height be?

Explanation:

1.

Step1: Recall vertex - form of a parabola

The equation of a parabola is \(y = ax^{2}+bx + c\), and the x - coordinate of the vertex is \(x=-\frac{b}{2a}\). For \(y = 2x^{2}-8x + 5\), where \(a = 2\), \(b=-8\), and \(c = 5\).
The x - coordinate of the vertex \(x=-\frac{-8}{2\times2}=\frac{8}{4}=2\).

Step2: Find the y - coordinate of the vertex

Substitute \(x = 2\) into \(y = 2x^{2}-8x + 5\).
\(y=2\times2^{2}-8\times2 + 5=2\times4-16 + 5=8-16 + 5=-3\).

Step1: Identify the coefficients

The height - time function is \(h=-16t^{2}+64t + 80\), where \(a=-16\), \(b = 64\), \(c = 80\).

Step2: Find the time \(t\) at the vertex

The time \(t\) at which the ball reaches its maximum height is given by \(t=-\frac{b}{2a}=-\frac{64}{2\times(-16)}=\frac{64}{32}=2\) seconds.

Step3: Find the maximum height

Substitute \(t = 2\) into \(h=-16t^{2}+64t + 80\).
\(h=-16\times2^{2}+64\times2+80=-16\times4 + 128+80=-64 + 128+80=144\) feet.

Step1: Identify the coefficients

The height - time function is \(h=-4.9t^{2}+27.7t\), where \(a=-4.9\), \(b = 27.7\), \(c = 0\).

Step2: Find the time \(t\) at the vertex

The time \(t\) at which the ball reaches the ground is when \(h = 0\). So, \(-4.9t^{2}+27.7t=0\). Factor out \(t\): \(t(-4.9t + 27.7)=0\). One solution is \(t = 0\) (corresponds to the time of throwing). The other solution is \(-4.9t+27.7 = 0\), then \(4.9t=27.7\), and \(t=\frac{27.7}{4.9}\approx5.65\) seconds.

Answer:

d. \((2,-3)\)

2.