McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
1. Graphing Quadratic Functions
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Exercise 58 Page 551

Practice makes perfect
a Miranda moves around the cape in the shape of a parabola. To graph the function d=-16t^2+66t, we need to plot the axis of symmetry, then the vertex and a couple more points.

Axis of Symmetry

From the function, d=-16t^2+66t, a = -16 and b=66. Let's substitute these into the formula for the axis of symmetry.
t=- b/2a
t=- 66/2( -16)
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Simplify right-hand side
t=- 66/-32
t= - -33/16
t=33/16
Let's plot the axis of symmetry on our coordinate plane.
Now, let's use that t-value to find d-value of the vertex.
d=-16t^2+66t
d=-16 ( 33/16)^2+66( 33/16)
â–Ľ
Simplify right-hand side
d= -16 ( 1089/256 )+ 66 ( 33/16 )
d= -17 424/256 + 2178/16
d= -1089/16+ 2178/16
d= 1089/16
Let's plot the vertex ( 3316, 108916)=(2.0625,68.0625).
We need two more points to get a complete picture of this parabola. Let's evaluate d when t=0 to find the d-intercept.
d=-16t^2+66t
d=-16( 0)^2+66( 0)
â–Ľ
Evaluate
d=-16(0)+66(0)
d=0+0
d=0
We now can plot the point (0,0) and it's mirrored point across the axis of symmetry. To find the mirrored point we use the same y-value and double the distance to the axis of symmetry for the x-value. We get ( 6616,0)=(4.125,0).

Now, we can connect the points with a parabola and finish the graph.


b To find the maximum number of feet north that Miranda traveled, we need to use the d-value of the vertex. In Part A, we found the vertex to be ( 3316, 108916) or (2.0625,68.0625). Therefore, she traveled 68.0625 feet north before starting southwards.
c To determine how long it took Miranda to get to her destination, we need to find where the curve crosses the t-axis. Fortunately, in Part A, we found both places the function intersects the t-axis. The second one is the point of Casper Marina, ( 6616,0)=(4.125,0). The t-value is the time it takes to get to that point, 4.125 minutes.