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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

 Jul 28, 2020
 #1
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+2

nce the line of symmetry is x = 2, the x coordinate of the vertex  must = 2

 

And the x coordinate of the vertex is given by :   -b/2a.......so we have that

 

2  = -b/[2a]      →   -4a = b

 

And we have these  equations

 

1 =  a(1)^2   - 4a(1)  + c    →   1 =  -3a + c       (1)

-1 = a(4)^2   - 4a(4)  + c    →  -1 =  c      (2)

 

Subbing (2)  into (1)  we have that    1 =  -3a - 1   →  2 = -3a →  a = -2/3

 

Which means that b  = -4(-2/3)  = 8/3

 

So....the equation of the quadratic is :

 

y = (-2/3)x^2 + (8/3)x  - 1

 

And we know that

 

0 = (-2/3)(√n +2)^2  + (8/3)(√n + 2)  - 1          let √n + 2   = m.......and we have that

 

0  = (-2/3)m^2 + (8/3)m  - 1      multiply both sides by 3

 

0 = -2m^2  + 8m - 3     complete the square on m

 

3 - 8 = -2 (m^2 - 4m + 4)

 

-5 = -2 (m - 2)^2

 

5/2  = (m - 2)^2       take pos/neg square roots

 

±√[5/2]  = m - 2

 

±√[5/2] + 2  = m     back-substituting, we have

 

±√[5/2] + 2 = √n + 2   and since we want the larger root

 

√[5/2]   = √n

 

n = 5/2  = 2.5

 

Here's the graph :  https://www.desmos.com/calculator/102umpw19h

 

The larger root = ( √2.5  + 2 ,  0 )  ≈  ( 3.581, 0 )

 Jul 28, 2020
 #3
avatar+26845 
+1

Close...but the vertex is in the wrong place....

    here is a graph:

https://www.desmos.com/calculator/atsyq7hk8i

ElectricPavlov  Jul 28, 2020
 #2
avatar+26845 
+3

In vertex form:    y = a (x-2)^2 + 4

Sub in the point given:

 

1 = a(1-2)^2 + 4

1 = a + 4                    a = -3

 

y = -3(x-2)^2 + 4       expand

y = -3x^2 +12x-8     

 Jul 28, 2020
 #4
avatar+254 
+1

Thanks! :-D

 

Nice job, EP 

 Jul 28, 2020

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