Question 479248: a door is parabolic in shape and its vertex is 8 feet above the floor. The width of the door way at floor level is 10 feet accordingly, would a 6 foot tall person hit their head, if the person walk through the doorway 2 feet from its center
Answer by solver91311(24713)   (Show Source):
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Set your parabola so that the vertex is at (0,8). That means the floor is the x-axis and the two intercepts are (-5,0) and (5,0).
Let the function  that describes this parabola be defined as \ =\ ax^2\ +\ bx\ +\ c)
Substituting the coordinate values from the three points we know must be on the parabola:
\ +\ b(0)\ +\ c\ =\ 8)
\ +\ b(-5)\ +\ c\ =\ 0)
\ +\ b(5)\ +\ c\ =\ 0)
Simplifying the first equation gives us:
Substituting and simplifying the other two equations results in the following 2X2 linear system:
The solution of the system and therefore the coordinates of the parabolic function is  ,  , and  , verification of which I leave as an exercise for the student.
Therefore the desired function is:
\ =\ -0.32x^2\ +\ 8)
Now if a six-foot tall person walks through the door so that the centerline of the person's body is 2 feet right or left of the centerline of the doorway, does the person hit their head?
If the person's head were infinitely thin, then it would be a simple matter of determining the value of the function at either  . However, in practical terms, given that the average six-foot tall person has significantly greater cranial volume than the average algebra student, we'll use 6 inches as a rough approximation of the width of the head, half of which is 3 inches. Hence, we need to determine the function value at 
Again, leaving the verification arithmetic to the student, a 6-foot tall person with a head the shape of a rectangular prism with a width along an axis parallel to the plane of the door of 6 inches would clear the door 2 feet either side of center by approximately 4.5 inches.
John
My calculator said it, I believe it, that settles it
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