How do you write the standard form of the equation of the parabola that has (-1/4, 3/2) vertex and (-2,0) point?

1 Answer
Sep 15, 2016

Answer:

#y=-24/49(x+1/4)^2+3/2# or #y=(-24/49)x^2-(12/49)x+72/49#

Explanation:

Use the equation #y=a(x-h)^2+k# where #(h,k)# is the vertex and a is a constant.

The given vertex is #(-1/4,3/2)#.
Plugging this into the equation above gives

#y=a(x--1/4)^2+3/2#

#y=a(x+1/4)^2 +3/2# (EQUATION A)

To find the constant #a#, substitute the given point #(-2,0)# in for #x# and #y#.

#0=a(-2+1/4)^2+3/2#

#0=a(-7/4)^2 +3/2#
#0=a(49/16)+3/2#
#0-3/2=a(49/16)#

#16/49*-3/2=a(49/16)*16/49#
#-24/49=a#

Substitute #a# into equation A above.
#y=a(x+1/4)^2+3/2#
#y=-24/49(x+1/4)^2+3/2#

Some books and teachers call this the standard form; but most call it the vertex form. To find the "usual" standard form, square the binomial and simplify.
#y=-24/49(x+1/4)(x+1/4)+3/2#
#y=-24/49(x^2+1/2x+1/16)+3/2#
#y=(-24/49)x^2 -(12/49)x-3/98+3/2#
#y=(-24/49)x^2-(12/49)x+72/49#