Finding the vertex of a parabola

[Anonymous No.8606068]

The variables [math]t,k[/math] are the coordinates [math](t,k)[/math] of the vertex.
Finding [math]k[/math] I can find [math]t[/math].
I wanted to find [math]t[/math] first, translating the parabola and imposing it to be symmetric.
It doesn't seem to work though.

[Anonymous No.8606081]

>>8606068
don understand english and maths sry

[Anonymous No.8606097]

Yeah this doesn't make a whole lot of sense.

Try drawing a picture?

[Anonymous No.8606153]

>>8606097
>Yeah this doesn't make a whole lot of sense.
It does.
I already found [math]k[/math].
I have and idea for finding [math]t[/math].

[math]x^{2}+2x+4=(x+2)^{2}[/math] is the translated [math]x^{2}[/math] parabola by [math]x\mapsto x+2[/math].
Thus if I am given [math]x^{2}+2x+4[/math], [math]x\mapsto x-2[/math] makes the graph symmetric and gives the x-coordinate of the vertex: [math]2[/math].
I tried to find [math]x\mapsto x+t[/math] that makes the graph symmetric imposing [math]y(x+t)=y(-x-t)[/math].

https://www.wolframalpha.com/input/?i=plot+(x%2B2)%5E2

[Anonymous No.8606170]

>>8606153
2 is not the x-coordinate of the vertex

[Anonymous No.8606222]

*[math](x+2)^{2}=x^{2}+4x+2[/math].
>>8606170
>2 is not the x-coordinate of the vertex
Yeah, you're right.
[math]t=-2[/math] in my example so I guess I missed a minus.
I see that you can rewrite the parabola as [math]y=a(x-t)^{2}+k[/math].
Still I'd like to know If I can find [math]t[/math] from the idea in
>>8606153.

[Anonymous No.8606235]

>>8606222
[math]t = -\frac{b}{2a}[/math]
Like, what else do you want?

[Anonymous No.8606267]

>>8606235
>t=−b2a
You don't say?
It's [math]y(f_{t}(x))=y(f_{t}(-x))[/math] the right condition and it works.
Imposing [math](x+t-2)^{2}=(-x+t-2)^{2}[/math] I find [math]t=2[/math].
Indeed [math]x\mapsto x-2[/math] makes [math](x+2)^{2}[/math] symmetric.
I only need to have a better look at what I wrote in order to keep the sign of [math]t[/math] is consistent.

[Anonymous No.8606270]

>>8606267
Fucking hell.