Question

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

Answer 1

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`