How do you write the equation of the parabola in standard form #x^2-12x-8y+20=0#?

1 Answer
Jun 18, 2016

Answer:

#y=1/8x^2-3/2x+5/2#

Explanation:

The standard form of a parabola is:
#y=ax^2+bx+c#

To find standard form, we must get #y# by itself on one side of the equation and all the #x#s and constants on the other side.

In order to do this for #x^2-12x-8y+20=0#, we must add #8y# to both sides, to get:
#8y=x^2-12x+20#

Then we must divide by #8# (which is the same thing as multiplying by #1/8#) to get #y# by itself:
#y=1/8x^2-3/2x+5/2#

The graph of this function is shown below.
graph{x^2-12x-8y+20=0 [-4.62, 15.38, -4.36, 5.64]}
#---------------------#

Bonus
Another common way of writing a parabola is in vertex form:
#y=a(x-h)^2+k#

In this form, #(h,k)# is the vertex of a parabola. If we write parabolas in this form, we can therefore easily identify the vertex, simply by looking at the equation (something we can't do with standard form).

The tricky part is getting it into this form, which often involves completing the square.

We'll start with the equation #8y=x^2-12x+20#, which is the same as #x^2-12x-8y+20=0# except with the #8y# in a different spot. We must now complete the square on the left side of the equation:
#8y=x^2-12x+20#
#8y=x^2-12x+36-16#
#8y=(x-6)^2-16#

Finish off by dividing by #8#, as we did previously:
#y=1/8(x-6)^2-2#

We can now instantly identify the vertex as #(6,-2)#, which can be confirmed by looking at the graph. (Notice that the #x#-point is #6# and not #-6# - it is easy to make that mistake). Using this fact, plus the #1/8# multiplier on #(x-6)^2#, we can gain a deeper understanding of the shape of the graph without even looking at it.