40 40 Maximum height 30 30 Shot Put's Height (feet) 20 20 g(x) = −0.04x² + 2.1.x + 6.1 Shot released at 65° Maximum height Distance of throw or maximum horizontal distance 10 f(x)=0.01x2 +0.7x+6.1 10 10 Shot released at 35° 90 20 30 40 50 60 70 80 90 Shot Put's Horizontal Distance (feet) x
40 40 Maximum height 30 30 Shot Put's Height (feet) 20 20 g(x) = −0.04x² + 2.1.x + 6.1 Shot released at 65° Maximum height Distance of throw or maximum horizontal distance 10 f(x)=0.01x2 +0.7x+6.1 10 10 Shot released at 35° 90 20 30 40 50 60 70 80 90 Shot Put's Horizontal Distance (feet) x
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
When the shot whose path is shown by the red graph on the
previous page is released at an angle of 65°, its height, g(x), in
feet, can be modeled by
g(x) =-0.04x
2
+2.1x+6.1,
where x is the shot’s horizontal distance, in feet, from its point
of release. Use this model to solve parts (a) through (c) and
verify your answers using the red graph.
a. What is the maximum height, to the nearest tenth of a
foot, of the shot and how far from its point of release does
this occur?
b. What is the shot’s maximum horizontal distance,
to the nearest tenth of a foot, or the distance of the
throw?
c. From what height was the shot released?
Transcribed Image Text:40
40
Maximum height
30
30
Shot Put's Height (feet)
20
20
g(x) = −0.04x² + 2.1.x + 6.1
Shot released at 65°
Maximum height
Distance of throw or maximum
horizontal distance
10
f(x)=0.01x2 +0.7x+6.1
10
10
Shot released at 35°
90
20 30 40 50 60 70 80 90
Shot Put's Horizontal Distance (feet)
x
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