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A quadratic passes through the points (-2,0) and (4,0), and has a maximum value of 54.  Find the equation of the quadratic.

 Dec 19, 2019
 #1
avatar+19789 
0

OK    ax^2 + bx + c = 0      the max is when x = 1  (the midpoint between the two given points)

  When x = 1   f(x) = 54     a(1^2) + b + c = 54      a + b + c = 54    c = 54-a-b

Sub in the values given

 

a (-2^2) + b(-2) + c = 0      4a-2b+c=0           4a-2b + 54-a-b = 0        3a-3b+54 = 0

a(4^2) + b(4) + c = 0       16a+4b+c = 0         16a+4b + 54 -a-b=0       15a+3b+54=0         Multiply green equation by -5 and add to this

 

-15a+15b-270

15a+3b+54

18b -216 = 0    b= 12     a =  -6     c =48

-6x^2 +12x + 48 = 0

 Dec 19, 2019
 #2
avatar+19789 
0

Here is a confirmation graph:

 

https://www.desmos.com/calculator/120umkmcim

ElectricPavlov  Dec 19, 2019
 #3
avatar+106519 
+2

(-2,0)    and  ( 4,0)

 

These must be the roots of the quadratic...so we have that

 

y  = a ( x + 2) ( x - 4)      simplify

 

y  = a (x^2 - 2x - 8)

 

And since  54 is a max, then using symmetry, the point  ( [-2 + 4]/2, 54)  = (1,54)  must be on the graph

 

So

54   =  a ( 1^2 - 2(1) - 8)

54 = a (-9)

a = -54/9  = -6

 

So.....the function is

 

y = (-6) ( x^2 -2x  - 8)

 

y = -6x^2 + 12x  + 48

 

Here's a graph  :  https://www.desmos.com/calculator/s8dqi0vdcn

 

 

cool cool cool

 Dec 19, 2019
 #4
avatar+19789 
0

Ugh.... this easier method escaped my thought !  Thanx !   ~EP

ElectricPavlov  Dec 19, 2019
 #5
avatar+106519 
0

Nothing wrong with your method, EP......just shows another way to solve it

 

This is always good   !!!!

 

 

cool cool cool

CPhill  Dec 19, 2019

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