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

Question

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

Eva Abramson · Accepted Answer

Vertex form: #y=a(x-h)^2+k# (In this textbook, vertex form is what it is called; other forms are also possible.) Method 1 It's likely that your teacher wants you to finish the square. (It's a useful trick that can be applied to other things too.) 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. This is how it appears: #y=3x^2-18x+5=3(x^2-6x " --------" ) +5# #=3(x^2-6x+9-9)+5#   Now reorganize, maintaining the ideal square where we desire it to be. (We are in control, within reason.) #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.) We'll write it like this now: #y=3(x-3)^2-27+5# #y=3(x-3)^2-22# That concludes our response. (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 Requires that you are familiar with the vertex formula, which was "found" by finishing 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#

Kennedy Adams · Answer

undefined To write the quadratic function $ y = 3x^2 - 18x + 5 $ in vertex form, follow these steps:

1. Factor out the coefficient of the $ x^2 $ term from the first two terms.
2. Complete the square for the $ x $ terms.
3. Rewrite the function in vertex form.

The vertex form of the quadratic function $ y = ax^2 + bx + c $ is $ y = a(x - h)^2 + k $, where $ (h, k) $ represents the coordinates of the vertex.

Let's apply these steps:

1. Factor out the coefficient of the $ x^2 $ term:

$ y = 3(x^2 - 6x) + 5 $

2. Complete the square for the $ x $ terms by adding and subtracting $ (\frac{b}{2})^2 $, where $ b = -6 $:

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

3. Rewrite the function in vertex form by factoring the perfect square trinomial and combining constants:

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

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

Therefore, the quadratic function $ y = 3x^2 - 18x + 5 $ in vertex form is $ y = 3(x - 3)^2 - 22 $.