How do you write a quadratic function in standard form whose graph passes through points (2,6), (-2,-2), (1,1)?

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Answer 1 · 2017-03-24T17:18:36

#y=x^2+2x-2#

Let the equation be #y=ax^2+bx+c#.

As it passes through #(2,6)#, #(-2,-2)# and #(1,1)#

#6=2^2a+2b+c# i.e. #4a+2b+c=6# ...(1)

and similarly #4a-2b+c=-2# ...(2)

and #a+b+c=1# ...(3)

Subtracting (2) from (1), we get #4b=8# i.e. #b=2#

this gives us #4a+c=2# and #a+c=-1#

and subtracting latter from former #3a=3# and #a=1#

which gives us #c=-2#

and our quadratic function is #y=x^2+2x-2#
graph{x^2+2x-2 [-10.83, 9.17, -3.4, 6.6]}

Answer 2 · 2017-03-24T17:19:22

#:. y=f(x)=x^2+2x-2, x in RR,# is the desired Quadr.

Let the Reqd. Quadr. Fun. be, #y=f(x)=ax^2+bx+c; a(ne0),b,c in RR.#

#(2,6) in f rArr 4a+2b+c=6............(1).#

#(-2,-2) in f rArr 4a-2b+c=-2...............(2).#

#(1,1) in f rArr a+b+c=1..........................(3).#

Solving these, #a=1, b=2, and, c=-2.#

#:. y=f(x)=x^2+2x-2, x in RR,# is the desired Quadr.