How do you write #y=3x^2-18x+5# in vertex form?

1 Answer
Mar 27, 2015

Vertex form: #y=a(x-h)^2+k#

(That's what vertex form is in the textbook here. Other forms are possible.)

Method 1

Your teacher probably wants you to complete the square. (It is an important technique, used for other things also.)

This method relies on the fact that: #(x+-a)^2=x^2+- (2a)x+a^2#

#y=3x^2-18x+5=3(x^2-6x " --------" ) +5# (Leave yourself some apce inside the parentheses)

If the stuff in parentheses is going to be a perfect square #(x-a)^2#, then it must have #2a=6# which tells me that #a=(1/2)(6)=3#.

So the stuff inside parentheses is not a perfect square becuase it is missing #a^2# which we would like to be #9#.

Now, (within limits) we are in charge here! We will add #9# inside the parentheses to make it a perfect square

#color(red) ("PROBLEM")# #3(x^2-6x+9)# is #3(x-3)^2#, but it is not equal to what we started with.

#color(green)("Solution")# We will add and subtract the #9# and regroup.

It looks like this:

#y=3x^2-18x+5=3(x^2-6x " --------" ) +5#

#=3(x^2-6x+9-9)+5#

Now regroup, keeping the perfect square were we want it to be. (We're in charge (within limits).)

#y=3(x^2-6x+9)-3(9)+5#
(Convince yourself that this really is equal to what we started with. Do the algebra to simplify it.)

Now we'll write it this way:

#y=3(x-3)^2-27+5#

#y=3(x-3)^2-22#

And there's our answer.

(If your class uses a different vertex for, it's probably:
#(y-k)=a(x-h)^2# so the answer would be #(y+22)=3(x-3)^2#)

Method 2

Depends on your knowing the vertex formula (which was 'discovered' by completing the square).

#y=ax^2+bx+c# has vertex with #x#-coordinate #(-b)/(2a)#

(Why> Well, start with #y=ax^2+bx+c# and complete the square to get #y=a(x+b/(2a))^2+"I'll leave it to you to find this"#

So #y=3x^2-18x+5# has vertex at #(-(-18))/(2(3))=3#

And when #x=3#,
we have #y=3(3)^2-18(3)+5=3(9)-6(9)+5=-3(9)+5=-27+5=-22#
(I now, I do arithmetic kinda weird.)
So #a=3#, #h=3# and #k=-22#

#y=3(x-3)^2-22#