I need help alg 2 | Mathematics | Page 1

mild orbitBOT Jul 23, 2025, 4:52 AM

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flat garnet Jul 23, 2025, 4:54 AM

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help ples

neat falcon Jul 23, 2025, 4:56 AM

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Well you're right that it is a parabola

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Are you familiar with completing the square?

flat garnet Jul 23, 2025, 4:59 AM

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nope

neat falcon Jul 23, 2025, 5:00 AM

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Were you given any formulas for computing the coordinates of the vertex of a parabola?

flat garnet Jul 23, 2025, 5:01 AM

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let me check

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this?

neat falcon Jul 23, 2025, 5:04 AM

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Hmm not exactly. This shows parabolas that have vertex at the origin (0,0).
The one in your problem is translated with a vertex at (h,k).

flat garnet Jul 23, 2025, 5:04 AM

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oh yeah yeah

neat falcon Jul 23, 2025, 5:04 AM

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Something like $h = \frac{-b}{2a}$

odd oxideBOT Jul 23, 2025, 5:04 AM

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Azyrashacorki

flat garnet Jul 23, 2025, 5:05 AM

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nope

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well i recorded the lesson earlier

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she solved it like this

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wait wrong pic

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well no

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i mean the way she teaches us kinda weird

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because when it comes to like homework and shit like everything is diff

neat falcon Jul 23, 2025, 5:08 AM

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Do you have formulas to compute the directrix and the focus then? Because here again the vertex is at (0,0).

flat garnet Jul 23, 2025, 5:08 AM

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nope

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sorry im really dumb when it comes to these things

neat falcon Jul 23, 2025, 5:09 AM

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It's not you being dumb I'm just trying to think of ways to compute the coordinates of the vertex the way your classes might've shown

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Once that's dealt with finding p is relatively easy.

flat garnet Jul 23, 2025, 5:10 AM

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do you wanna see the answer? because i can re do it

neat falcon Jul 23, 2025, 5:10 AM

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odd oxide **Azyrashacorki**

In any case this works

flat garnet Jul 23, 2025, 5:10 AM

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here

neat falcon Jul 23, 2025, 5:11 AM

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flat garnet do you wanna see the answer? because i can re do it

No it's ok I know how to do it, it's just there are multiple ways to go about it I don't want to show you stuff that will confuse you if you've not seen them in class.

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So if you have $y= ax^2 + bx + c$, the vertex is at $(h,k)$ where $h = \frac{-b}{2a}$ and $k = ah^2 = bh + c$.

odd oxideBOT Jul 23, 2025, 5:12 AM

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Azyrashacorki

neat falcon Jul 23, 2025, 5:12 AM

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And $p = \frac{1}{4a}$

odd oxideBOT Jul 23, 2025, 5:12 AM

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Azyrashacorki

flat garnet Jul 23, 2025, 5:13 AM

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i mean you can just show me how does it done

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and ill learn from it

neat falcon Jul 23, 2025, 5:19 AM

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Those formulas are how it's done tbh

Otherwise as I pointed out earlier there's square completion :
$3x^2 + 30x + 68 = 3(x^2 + 10x + \frac{68}{3})$.

Now $$x^2 + 10x = (x+5)^2 - 25$$, so you get $$y = 3((x+5)^2 -25 + \frac{68}{3}) = 3((x+5)^2 - \frac{7}{3} = 3(x+5)^2 - 7.$$

Send the 7 to the other side and you get $y+7 = 3(x+5)^2$ and you have pretty much all the information for h, k and p.

odd oxideBOT Jul 23, 2025, 5:19 AM

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Azyrashacorki

flat garnet Jul 23, 2025, 5:19 AM

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BRUH

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AINT NO WAY THATS IT

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THANK YOU

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GOOD LUCK ON YO LEAGUE

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.done

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how do yo end this shit

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.done

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/done

neat falcon Jul 23, 2025, 5:27 AM

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.close

flat garnet Jul 23, 2025, 5:29 AM

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bruh

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tahnk you again

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.close

#I need help alg 2