What is the left-hand limit of this function as x approaches 2? f(x)=x/x-2
1 answer:
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The heights the balls hit a fence at 350 ft distance are 65 feet, 38 feet and 30 feet, respectively
<h3>Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. </h3><u>Juan</u>
Juan's equation is given as:
h = -0.001d^2 + 0.5d + 2.5
h =
Set d to multiples of 50 from 0 to 400.
So, the table of values of Juan's function is:
d (ft) h(ft)
0 2.5
50 25
100 42.5
150 55
200 62.5
250 65
300 62.5
350 65
400 42.5
See attachment for the graph of Juan's function
<u>Mark</u>
A quadratic function is represented as:
h = ad^2 + bd + c
Using the values on the table of values, we have:
c = 3 -- the constant value
So, the equation becomes
h = ad^2 + bd + 3
Using the two other values on the table of values, we have:
23 = a(50)^2 + b(50) + 3
38 = a(100)^2 + b(100) + 3
Using a graphing tool, we have:
a = -0.001
b = 0.45
So, Mark's equation is h(d) = -0.001d^2 + 0.45d + 3
See attachment for Mark's graph.
<u>Barry</u>
From the graph, we have the table of values of Barry's function to be:
d (ft) h(ft)
0 2.5
50 21
100 35
150 44
200 48
250 46
300 41
350 30
400 14
450 0
Using a graphing tool, we have the quadratic function to be:
y = -0.001x^2 +0.4x +2.5
<h3><u>The shortest and the greatest distance before hitting the ground</u></h3>From the graphs, equations and tables, the distance travelled by the balls are:
Juan = 505 feet
Mark = 457 feet
Barry = 450 feet
This means that Juan's ball would travel the greatest distance while Barry's ball would travel the shortest.
<h3>The height the balls hit a fence at 350 ft distance</h3>To do this, we set d = 350
From the graphs, equations and tables, the height at 350 ft by the balls are:
Juan = 65 feet
Mark = 38 feet
Barry = 30 feet
The above represents the height the balls hit the fence
Read more about quadratic functions at:
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