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Find a polynomial function whose graph passes through (6,13), (9,- 12), and (0,4)

 Jan 8, 2020
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Well....we can generate a quadratic for this

 

Let the  polynomial be

 

y = ax^2 + bx + c

 

Since  (0, 4)   is in the graph, then  c  = 4

 

And using the other two points we  have this system

 

13  = a(6)^2  + b(6) + 4

 -12 = a(9)^2  + b(9) + 4        simplify

 

36a + 6b  = 9

81a + 9b  = - 16         

 

Divide the   first equation through by 6

6a + b = 3/2

b = (3/2) - 6a          sub this into the second equation for  b   and we have that

 

81a +9(3/2 -6a)  = -16

81a + 27/2 - 54a   = -16

27a  =    -16 - 27/2

27a  =  [ -32 - 27 ] / 2

27a  = [ -59 ] /2    

a =  [ -59/27] / 2

a = -59/54

 

And

b  =  (3/2)  - 6 (-59/54 )

b = (3/2) + 59/9

b = 145/18

 

So.....the polynomial is

 

-(59/54)x^2  + (145/18)x +  4 

 

Here's a graph  :  https://www.desmos.com/calculator/jrc6i4llx4

 

 

cool cool cool

 Jan 8, 2020

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