How do you write the standard form of the equation of the parabola that has (3,4) vertex and (1,2) point?

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Answer 1 · 2017-03-07T08:05:53

Equation is #y=-x^2/2+3x-1/2# or #x=-y^2/2+4y-5#

As the vertex is #(3,4)#, the equation in vertex form could be either #y=a(x-3)^2+4# or #x=a(y-4)^2+3#

If it is #y=a(x-3)^2+4# and passes through #(1,2)#, we have

#2=a(1-3)^2+4# or #2=4a+4# i.e. #a=-1/2#

and equation of parabola is #y=-(x-3)^2/2+4#

or #y=-x^2/2+3x-9/2+4# i.e. #y=-x^2/2+3x-1/2#

If equation is considered to be #x=a(y-4)^2+3# and passes through #(1,2)#, we have

#1=a(2-4)^2+3# or #1=4a+3# i.e. #a=-1/2#

and equation of parabola is #x=-(y-4)^2/2+3#

or #x=-y^2/2+4y-8+3# i.e. #x=-y^2/2+4y-5#

and graphs of thetwo equation appear as follows:
graph{(y+x^2/2-3x+1/2)(x+y^2/2-4y+5)=0 [-8.79, 11.21, -2.56, 7.44]}