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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". Thanks!

 Dec 30, 2019
 #1
avatar+19806 
0

Here is a way:

 

Vertex value of 'x' is at   -b/2a     soooo     -b/2a = 2  (1) 

                                                  and when x = 2        f(x)=4

                                                            4 = a(2^2)+b(2) + c   or  4 = 4a+2b+c         (2)

    and you are given point 1,1            1 = a(1^2)+b(1)+c     or   1 = a+b+c            (3)       3 equations , 3 unknowns

 

re-arrange (1) :   -b/2 = 2a      multiply by 2

                               -b =4a          sub into (2)

 4 = -b+2b+c    4 = b+c     c = 4-b

 

Take (3) and multiply by 4     4 = 4a + 4b + 4c  

                                             4 = -b+4b + 4(4-b)    so b = 12    then     -b/2a=2 means   a = -3      then a + b + c =1 means c = -8

 Dec 30, 2019
 #2
avatar+139 
0

Wow thanks a lot Eletric Pavlov you are very smart

 Dec 30, 2019
 #3
avatar+11146 
+1

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". Thanks!

laugh

 Dec 30, 2019

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