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A parabola ax^2+bx+c contains the points (-1,0), (0,5), and (5,0). Find the value of 100a + 10b + c.

Guest Feb 3, 2018

Best Answer 

 #1
avatar+27219 
+2

(-1,5):   a(-1)2+b(-1)+c = 0 or a - b + c = 0

 

(0,5):    a(0)2+b(0)+c = 5 or c = 5

 

(5,0):   a(5)2+b(5)+c = 0 or 25a + 5b +c = 0

 

So: 

a - b + 5 = 0.        (1)

25a + 5b + 5 = 0. (2) 

 

Multiply (1) by 5 and add to (2):

30a + 30 = 0.  So a = -1 

 

Hence, From (1): 

-1 - b + 5 = 0. So b = 4

 

a = -1, b = 4, c = 5

 

I’ll leave you to calculate 100a + 10b + c

Alan  Feb 3, 2018
edited by Alan  Feb 3, 2018
 #1
avatar+27219 
+2
Best Answer

(-1,5):   a(-1)2+b(-1)+c = 0 or a - b + c = 0

 

(0,5):    a(0)2+b(0)+c = 5 or c = 5

 

(5,0):   a(5)2+b(5)+c = 0 or 25a + 5b +c = 0

 

So: 

a - b + 5 = 0.        (1)

25a + 5b + 5 = 0. (2) 

 

Multiply (1) by 5 and add to (2):

30a + 30 = 0.  So a = -1 

 

Hence, From (1): 

-1 - b + 5 = 0. So b = 4

 

a = -1, b = 4, c = 5

 

I’ll leave you to calculate 100a + 10b + c

Alan  Feb 3, 2018
edited by Alan  Feb 3, 2018

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