Find quadratic function?

Question

I'm having trouble with the following problem.  If anybody can explain how to set up the quadratic, I'd appreciate it.
Thanks!

You are 5.5 feet tall and are shooting a free throw.  The path  of the ball is parabolic in shape and reaches its maximum height of 11.5 feet when the ball is 10 feet from you.  Find the function for the path of the ball.  Let x be the horixontal distance from you, and y be the height of the ball above the ground.

gravitative · Accepted Answer

Since it gives us the highest point / maximum / vertex:

Vertex form:
ƒ(x) = a·( x - h )² + k

Where h is the horizontal shift (x value of the vertex)
and k is the vertical shift (y value of the vertex)

Since "x [= horizontal] distance from you"
and "y [=] height ... above the ground"
Plug in h = 10 and k = 11.5
ƒ(x) = a·[ x - (10) ]² + (11.5)
ƒ(x) = a·( x - 10 )² + 11.5

We still need a though. Fortunately, we have another point:
"5.5 feet tall and are shooting a free throw"
Height above ground (y) is 5.5 when horizontal distance (x) is 0.

Plug the point (0,5.5) into the equation and solve for a:

5.5 = a·( 0 - 10 )² + 11.5
5.5 = a·( -10 )² + 11.5
5.5 = a·100 + 11.5
5.5 - 11.5 = a·100
-6 = a·100
-6/100 = a
-3/50 = a

----------------
Answer
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y = -3/50·( x - 10 )² + 11.5


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Alternative method
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It's possible but excessively difficult without vertex form:

Solving with standard form:
ƒ(x) = A·x² + B·x + C

Plug in (10,11.5):
11.5 = A·(10)² + B·(10) + C
11.5 = A·(100) + B·(10) + C
- C + 11.5 = A·(100) + B·(10)
- C = 100·A + 10·B - 11.5
C = - 100·A - 10·B + 11.5

Substitute:
ƒ(x) = A·x² + B·x - 100·A - 10·B + 11.5

Plug in (0,5.5) to this new formula:
ƒ(x) = A·x² + B·x - 100·A - 10·B + 11.5
5.5 = A·(0)² + B·(0) - 100·A - 10·B + 11.5
5.5 = - 100·A - 10·B + 11.5
10·B + 5.5 = - 100·A  + 11.5
10·B = - 100·A  + 11.5 - 5.5
10·B = - 100·A  + 6
B = -10·A  + 0.6

Substitute for B:
ƒ(x) = A·x² + (-10·A  + 0.6)·x - 100·A - 10·(-10·A  + 0.6) + 11.5
ƒ(x) = A·x² + (-10·A  + 0.6)·x - 100·A + 100·A  - 6 + 11.5
ƒ(x) = A·x² + (-10·A  + 0.6)·x + 5.5


This method needs a third point.
To get a third point, recognize that (10,11.5) is the vertex, so points on the left and right of it are symmetric.  Since (0,5.5) is 10 units to the left of the vertex, there is an identical point 10 units to the right of the vertex at (20,5.5).

Plug in (20,5.5):
ƒ(x) = A·x² + (-10·A  + 0.6)·x + 5.5
5.5 = A·(20)² + (-10·A  + 0.6)·(20) + 5.5
5.5 = A·(400) + (-200·A  + 12) + 5.5
5.5 = 200·A  + 17.5
5.5 - 17.5 = 200·A
-12/200 = A
-0.06 = A

-0.06 = A
B = -10·(-0.06)  + 0.6
B = 0.6  + 0.6
B = 1.2
C = - 100·(-0.06) - 10·(1.2) + 11.5
C = 6 - 12 + 11.5
C = 5.5

ƒ(x) = -0.06·x² + 1.2·x + 5.5

... so I'd really recommend the first method.

Find quadratic function?

1 Answer