How do you find the standard form of the equation for the parabola with vertex (4,0) and passing through the point (1,2)?

1 Answer
Shwetank Mauria
May 14, 2018

Equation is #y=2/9x^2-16/9x+32/9# or #x=-3/4y^2+4#

Explanation:

With vertex at #(h,k)#, we can have two types of parabolas - (i) of regular vertical type #y=a(x-h)^2+k# and (ii) of horizontal type #x=a(y-k)^2+h#. In standard form it is #y=ax^2+bx+c# or #x=ay^2+by+c#

If it is regular type, if vertex is #(4,0)#, our equation is

#y=a(x-4)^2+0# and as it passes through #(1,2)#, we have

#2=a(1-4)^2# or #9a=2# i.e. #a=2/9# and

equation is #y=2/9(x-4)^2# or #9y=2(x-4)^2#

or in standard form equation is #y=2/9x^2-16/9x+32/9#

If parabola is horizontal, if vertex is #(4,0)#, our equation is

#x=a(y-0)^2+4# and as it passes through #(1,2)#, we have

#1=a*2^2+4# or #4a=-3# i.e. #a=-3/4# and

equation is #x=-3/4y^2+4# or #4x+3y^2=16#

and in standard form equation is #x=-3/4y^2+4#
graph{(4x+3y^2-16)(2(x-4)^2-9y)=0 [-10, 10, -5, 5]}

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
4029 views around the world
You can reuse this answer
Creative Commons License