Question

What would be the vertex from this vertex form?

-(x - 2)^2 - 5

Answer 1

@jim_thompson5910

Answer 2

@Nnesha

Answer 3

you first need to get it into the form \[\Large a(x-h)^2 + k\]

Answer 4

i know that vertex would be the h and k

would this be (-2, -5) or (2, -5)

Answer 5

oh nvm i misread, it's already in that form

Answer 6

\[\Large -(x - 2)^2 - 5\]
\[\Large -1(x - 2)^2 + (-5)\]
\[\Large \color{green}{-1}(x - \color{red}{2})^2 + (\color{blue}{-5})\]
Compare to
\[\Large \color{green}{a}(x - \color{red}{h})^2 + \color{blue}{k}\]
and we see that
\[
\Large \color{green}{a} = \color{green}{-1}\\
\Large \color{red}{h} = \color{red}{2}\\
\Large \color{blue}{k} = \color{blue}{-5}
\]
So the vertex is
\[\Large \left(\color{red}{h},\color{blue}{k}\right) = \left(\color{red}{2},\color{blue}{-5}\right)\]

Answer 7

we distribute the -1 to -2 to get 2

Answer 8

oh no!!

Answer 9

follow exactly what @jim_thompson5910 wrote above

Answer 10

you cannot distribute unless you multiply out first, i.e. square it
but you do not need to do that here, since your quadratic was already written in vertex form