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The graph of $y = ax^2 + bx + c$ has a maximum value of 54, and passes through the points $(-2,0)$ and $(4,0).$ Find $a + b + c.$

 Jun 25, 2019
 #1
avatar+103069 
+2

y = ax^2 + bx +  c

 

The vertex of the parabola  will  occur  at   [  (-2 + 4) / 2 ,  54 ]  =    [2/2 , 54]  = [ 1, 54]

 

And the x coordinate of the vertex  =   -b/ [2a ]  = 1     so   b = -2a

 

So we have this system of equations

 

a(1)^2  -2a(1) + c =  54    ⇒   -a + c  = 54  ⇒    a - c  = -54

a(-2)^2  -2a(-2) + c  = 0   ⇒  8a + c = 0  ⇒      8a + c   = 0     

 

Add the last two equations and we get that

 

9a  =  -54

a  = - 6

 

b = -2(-6)  = 12

 

And

 

8(-6) + c  = 0   ⇒  c = 48

 

So   a  + b  + c    =    -6 +  12  +  48  =    54

 

Here's the graph :  https://www.desmos.com/calculator/9u0epc5dbu

 

 

cool cool cool

 Jun 25, 2019
 #2
avatar+8720 
+4

Also since we know all the zeros are at   x = -2   and   x = 4 ,  we can say

 

y  =  a(x + 2)(x - 4)

 

Since the axis of symmetry on a parabola is always halfway between the roots, and since the maximum is 54....

 

vertex   =   ( \(\frac{4-2}{2}\), 54)   =   (1, 54)

 

Now that we know  (1, 54)  is a solution to  y = a(x + 2)(x - 4) , we can say

 

54  =  a(1 + 2)(1 - 4)

 

54  =  a( -9 )

 

-6  =  a

 

So the equation of the parabola is...

 

y  =  -6(x + 2)(x - 4)

 

y  =  -6( x2 - 2x - 8 )

 

y  =  -6x2 + 12x + 48

 

-6 + 12 + 48  =  54

 Jun 25, 2019
 #3
avatar+91 
+1

Thanks :)

 Jun 25, 2019

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