McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Standardized Test Practice

Exercise 14 Page 655

Practice makes perfect
a When given a graph of an equation, we can sometimes find the vertex by looking. In this case, to be precise, we need to find the axis of symmetry and the height where the curve meets the axis.
t=- b/2a
t = - 128/2( -16)
Evaluate right-hand side
t =- 128/-32
t = - (-4)
t = 4
Now, to find the h-value, we need to substitute t=4 into our function and evaluate.
h(t) = -16t^2 +128t
h( 4)= - 16( 4)^2 + 128( 4)
Evaluate right-hand side
h(4) = - 16(16) + 128(4)
h(4) = - 256 + 512
h(4) = 256
Therefore, the vertex is (4,256), which corresponds to the vertex in the graph of the function in the exercise.
b From Part A, we know the vertex is ( 4, 256). From the graph, we can see that the first coordinate, t, represents time in seconds and the second coordinate, h represents height in feet. Therefore we can say that the highest point of the firework is 256 feet, 4 seconds after it is launched.
c The axis of symmetry represents the halfway point of the launch. From Part B, we know it takes the firework 4seconds to reach the vertex from the ground. Therefore, it will take it the same amount of time to travel from the vertex to the ground, 4 * 2 or 8 seconds.