How do you write a quadratic function whose graph has the given characteristics: vertex (2,7), passes through (4,2)?

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Answer 1 · 2017-05-13T06:22:01

#5x^2-20x+4y-8=0#

or #2y^2-28y-25x+148=0#

Given vertex #(h,k)#, equation of quadratic function could be #y=a(x-h)^2+k# or #x=a(y-k)^2+h#.

As vertex is #(2,7)#, using #y=a(x-h)^2+k# gives us

#y=a(x-2)^2+7# and as it passes through #(4,2)#

we have #2=a(4-2)^2+7# or #4a=-5# i.e. #a=-5/4#

and equation is #y=-5/4(x-2)^2+7#

or #4y=-5x^2+20x-20+28# or #5x^2-20x+4y-8=0#

Using #x=a(y-k)^2+h#, we get #x=a(y-7)^2+2# and as it passes through #(4,2)#, we get

#4=a(2-7)^2+2# i.e. #a=2/25# and equation is

#x=2/25(y-7)^2+2# or #25x=2y^2-28y+98+50# or #2y^2-28y-25x+148=0#

graph{(5x^2-20x+4y-8)(2y^2-28y-25x+148)=0 [-9.84, 10.16, -2.04, 7.96]}