Question

Write this in vertex form F(x)=x^2+6x-4

Answer 1

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

Answer 2

yeah that is when the first and last term are squared

Answer 3

\(\bf f(x)=x^2+6x-4\quad \textit{let's start by grouping the "x"'s}\\ \quad \\
f(x)=(x^2+6x)-4\implies f(x)=(x^2+6x+{\color{red}{ \square }}^2)-4\)

what number there, would give us a perfect square trinomial?

Answer 4

would it be 9 or 3?

Answer 5

\(\bf f(x)=x^2+6x-4\quad \textit{let's start by grouping the "x"'s}\\ \quad \\
f(x)=(x^2+6x)-4\implies f(x)=(x^2+6x+{\color{red}{ 3 }}^2)-4\\ \quad \\\implies (x-{\color{red}{ 3}})^2-4{\color{red}{ -3^2}}\)

notice, all we're doing borrowing from zero, 0, so if we ADDED 9, or \(3^2\), we also have to SUBTRACT it

Answer 6

\(\bf f(x)= (x-{\color{red}{ 3}})^2-4{\color{red}{ -3^2}}\implies f(x)=(x-3)^2-13\)

Answer 7

ohh wait.. is +... dohh my bad

Answer 8

\(\large f(x)= (x+{\color{red}{ 3}})^2-4{\color{red}{ -3^2}}\implies
\begin{array}{llll}
f(x)=(x&+3)^2&-13\\
&(h,&\ k)\Leftarrow vertex
\end{array}\)

Answer 9

so f(x)=(x+3)^3-13 would the vertex form?

Answer 10

\(a\)