How do write standard form of equation of the parabola given that vertex is (5/2, - 3/4) and graph passes through point (-2 ,4)?

1 Answer
Aug 23, 2016

Answer:

Express the parabola in vertex form and work from there to obtain #y=19/81x^2-95/81x+58/81#.

Explanation:

We might just want to jump into the problem and try finding the standard form #y=ax^2+bx+c# of this parabola, but we don't have enough information to do this yet. Instead, we have to use the vertex form of a parabola:
#y=a(x-h)^2+k#
Where #(h,k)# is the vertex.

The vertex is given in the problem, so:
#y=a(x-5/2)^2-3/4#

We have to find the value of the constant #a# (which determines how thin or thick the parabola is), and we can do this using the point #(-2,4)#:
#y=a(x-5/2)^2-3/4#
#->4=a(-2-5/2)^2-3/4#
#->4+3/4=a(-9/2)^2#
#->19/4=81/4a#
#->a=19/81#

Our equation in vertex form is therefore:
#y=19/81(x-5/2)^2-3/4#

We can expand the right hand side to find the standard form:
#y=19/81(x-5/2)^2-3/4#
#->y=19/81(x^2-5x+25/4)-3/4#
#->y=19/81(x^2-5x+25/4)-3/4#
#->y=19/81x^2-95/81x+475/324-3/4#
#->y=19/81x^2-95/81x+58/81#