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

tertre  May 31, 2017
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3+0 Answers

 #1
avatar+75302 
+2

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

 

It's obvious that  c  =  5

 

So

 

a(-1)^2  + b(-1)  + 5   =  0   →    a -  b  + 5  =  0  →   a - b   =  -5  →  a + 5  =  b   (1)

 

And

 

a(5)^2  + b(5)  +  5    =   0    →   25a + 5b  =  -5   →   5a  + b  =  -1     (2)

 

Sub  (1)  into (2)

 

5a + (a + 5)  =  -1

 

6a + 5  = -1      subtract  5 from both sides      

 

6a  =  -6

 

a  =  -1          and      a +  5  =  b  ....so......   -1 + 5  =  b   = 4

 

So  our function  is

 

y =  -1x^2  + 4x  +  5

 

So

 

100a  +  10b +  c  =

 

100 (-1)  + 10 (4)  + 5  =

 

-100 +  40  + 5   =

 

-100 +  45  =

 

-55

 

 

cool cool cool

CPhill  May 31, 2017
edited by CPhill  May 31, 2017
 #2
avatar+4154 
+2

Also...since they gave us the points    (-1, 0)    and    (5, 0)

 

These are the roots

So we can say

 

y = a(x + 1)(x - 5)

 

And plug in the point (0,5) to find a.

5 = a(0 + 1)(0 - 5)

5 = a(-5)

-1 = a

 

So....

y = -(x + 1)(x - 5)

y = -(x2 - 4x - 5)

y = -x2 + 4x + 5

hectictar  May 31, 2017
 #3
avatar+75302 
+1

 

Thanks, hectictar.....!!!!!

 

 

 

cool cool cool

CPhill  May 31, 2017

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