SOLUTION: Can you help with this question? An athelete whose event is the shot put releases a shot. When the shot whose path is shown by the graph is releases at an angle of 45 degrees, it

Question 230225: Can you help with this question?
An athelete whose event is the shot put releases a shot. When the shot whose path is shown by the graph is releases at an angle of 45 degrees, its height, f(x), in feet, can be modeled by f(x)=-0.02x^2+1.0x+6.5, where x is the shots horizontal distance, in feet, from its point of release.
a. Where is the maximum height of the shot and how far from the point of release does this occur
b. What is the shots maximum horizontal distance to the nearest tenth, of a foot or the distance of the throw.
c. From what height was the shot released?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An athlete whose event is the shot put releases a shot.
When the shot whose path is shown by the graph is releases at an angle of 45
degrees, its height, f(x), in feet, can be modeled by f(x)=-0.02x^2+1.0x+6.5,
where x is the shots horizontal distance, in feet, from its point of release.
:

a. Where is the maximum height of the shot and how far from the point of release does this occur?
You can see on the graph that max height is about 20 ft and occurs at about 25 ft
however, you can calculate this: Find the axis of symmetry (x=-b/(2a)
In this equation a=-.02; b=1
x =
x =
x = +25 is the horizontal distance
Find the height: substitute 25 for x in the given equation
h = -0.02(25^2) +1.0(25)+ 6.5
h = -0.02(625) + (25)+ 6.5
h = -12.5 + 25 + 6.5
h = 19 ft is the max height
:
:
b. What is the shots maximum horizontal distance to the nearest tenth, of a foot or the distance of the throw.
That occurs when h = 0
.02x^2 + 1x + 6.5 = 0
Use the quadratic formula: a=-.02; b=1; c=6.5
The positive solution: x = 55.8 ft
;
c. From what height was the shot released?
The value of h when x = 0 is 6.5 ft (you can see this on the graph)
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