Question

Explain how to convert standard form to vertex form?


1) “To pass by me you must tell me how to convert standard form into the general, vertex form... I have a test on it next week.”

Explain how to convert f(x) into the general, vertex form of the equation. Use complete sentences.

f(x) = x^2 + 6x – 16

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

I think so. When you multiply a binomial by itself, right?

Answer 3

well, yes, but a perfect square trinomial has.... 3 terms that look like this \(\bf a^2\pm {\color{red}{ 2}}ab+b^2\) <-- notice the the middle term
the middle term is composed of the root of the two extremes times 2

Answer 4

so.. it ends up at -> \(\bf a^2\pm {\color{red}{ 2}}ab+b^2\implies (a\pm b)^2\)

Answer 5

Ok so I need to rearrange f(x) into (a plus or minus b)^2?

Answer 6

so let's take a peek at the equation, and lets firstly do some grouping \(\bf f(x) = x^2 + 6x - 16\implies f(x) = (x^2 + 6x) - 16
\\ \quad \\
\implies f(x) = (x^2 + 6x+{\color{red}{ \square }}^2) - 16\)

so what do you think we need there to make that a perfect square trinomial?

Answer 7

6x^2?

Answer 8

well.... notice the middle term, keep in mind the middle term is using a factor of "2"

if you remove that factor of 2, what would you be left?

Answer 9

3?

Answer 10

yeap, 3 thus 2*3*x = 6x

keep in mind that, all we're doing is borrowing from "0", zero
so if we ADD \(3^2\) we also have to SUBTRACT \(3^2\) thus

\(\bf f(x) = x^2 + 6x - 16\implies f(x) = (x^2 + 6x) - 16
\\ \quad \\
f(x) = (x^2 + 6x+{\color{red}{ 3 }}^2) - 16-{\color{red}{ 3 }}^2\implies f(x)= (x+3)^2-16-9
\\ \quad \\
\implies f(x)= (x+3)^2-25\)

Answer 11

So how would I put this in complete sentences?
f(x)=x^2+6x-16 can be converted to vertex form by grouping it to make a perfect square trinomial to get f(x)=(x^2+6x+3^2)-16-3^2 by adding 3^2 and also subtracting 3^2 to get f(x)=(x+3)^2-25 ?

Answer 12

yeap

Answer 13

Thank you!

Answer 14

yw