Question

Help? Will give fan and medal.

What is the vertex form of the equation?
y = -x^2 + 12x - 4

Answer 1

do you know what a "perfect trinomial" is?

Answer 2

I'm sorry! I got busy for a little bit. And although it sounds very familiar, I can't remember. @SolomonZelman

Answer 3

There is a formula;
\(\color{#000000 }{ \displaystyle (x+a)^2=x^2+2ax+a^2 }\)
And the right side is called a "perfect square trinomial"
because the right side is a perfect square, except that
it is a trinomial - not a number.

When you multiply \(\color{#000000 }{ \displaystyle (x+a)}\) times another \(\color{#000000 }{ \displaystyle (x+a) }\),
you get;
\(\color{#000000 }{ \displaystyle (x+a)\cdot (x+a)= }\)
\(\color{#000000 }{ \displaystyle x(x+a)+ a(x+a)= }\)
\(\color{#000000 }{ \displaystyle x^2+ax+ ax+a^2= }\)
\(\color{#000000 }{ \displaystyle x^2+2ax+a^2. }\)

(that proves the formula, and it is a perfect square. Just
as when you multiply 4•4 and get 16, 16 is a perfect square
because it's square root is an integer 4, SO IS \(\color{#000000 }{ \displaystyle x^2+2ax+a^2 }\)
a perfect square, because when you multiply (x+a) times (x+a)
you get \(\color{#000000 }{ \displaystyle x^2+2ax+a^2 }\))

So, we can infer that for a trinomial to be a "perfect square"
we have to put it in a form \(\color{#000000 }{ \displaystyle x^2+2ax+a^2 }\).

You might wonder why you need to know about it, but
the vertex form of the parabola looks like this:
\(\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\)

And notice we have \(\color{#000000 }{ \displaystyle (x-h)^2 }\) sitting in there.
Aha! This is why we need a perfect square trinomial,
because \(\color{#000000 }{ \displaystyle x^2+2ax+a^2 }\) is (same as) \(\color{#000000 }{ \displaystyle (x+a)^2 }\); that
way we can put our equation into the "vertex form".


\(\color{red}{_\text{___________________________________________________________________}}\)
More about vertex form eq. of parabola:
Form; \(\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\)
Vertex; \(\color{#000000 }{ \displaystyle (h,k) }\)
Opens down: \(\color{#000000 }{ \displaystyle a<0 }\) (negative a)
vertex is the maximum point
(maximum of y=k, that occurs at x=h)

Opens up: \(\color{#000000 }{ \displaystyle a>0 }\) (positive a)
vertex is the minimum point
(minimum of y=k, that occurs at x=h)
\(\color{red}{^\text{___________________________________________________________________}}\)

Answer 4

It makes sense :) thank you very much!

Answer 5

This was really helpful.

Answer 6

site glitches...

Answer 7

I am putting an example up

Answer 8

Site glitches?

Answer 9

I will try to give an example to clarify some things...

Suppose I want to write \(\color{#000000 }{ \displaystyle y=x^2-4x }\) in a vertex form.

Then, I will need to put my equation in a form of \(\color{#000000 }{ \displaystyle x^2+2ax+a^2 }\).
I already have the \(\color{#ff0000 }{ \displaystyle x^2+2ax }\) (and that is \(\color{#000000 }{ \displaystyle x^2-4x }\)), so what I need
to do, is to add "\(a^2\)" component, and that I will be able to do only
if I figure out the \(a\).

I can find \(a\) via comparison.
\(\color{#ff0000 }{ \displaystyle x^2+2ax }\) in the form, CORRESPONDS to \(\color{#ff0000 }{ \displaystyle x^2-4x }\) in our equation,
so, we can tell that \(2a=-4\) --> \(a=-2\). Therefore, "\(a^2\)" in our
case is going to be \(\color{#000000 }{ \displaystyle a^2\Longrightarrow (-2)^2=4 }\)

So we would need
\(\color{#000000 }{ \displaystyle y=x^2-4x\color{blue}{+4} }\)
but, we can NOT just stick a \(4\) in there (right?)

So, we will do a trick!
\(\color{#000000 }{ \displaystyle y=x^2-4x\color{blue}{+4}\color{red}{-4} }\)
(by adding 4 and subtracting 4, I am not changing the value)

Now, we can see the perfect square part underlined,
\(\color{#000000 }{ \displaystyle y=\underline{x^2-4x\color{blue}{+4}}\color{red}{-4} }\)
and this therefore becomes,
\(\color{#000000 }{ \displaystyle y=\underline{(x-2)^2}\color{red}{-4} }\)
(I have -2 in the parenthesis, because I already found a to be -2)

So \(\color{#000000 }{ \displaystyle y=x^2-4x }\)
is written as \(\color{#000000 }{ \displaystyle y=(x-2)^2-4 }\) in a vertex form.

Answer 10

(if you want another example or have questions.... u know....)