find y=a(x-h)^2+m | Mathematics | Page 1

sinful shard Feb 3, 2023, 2:08 AM

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find p y=a(x-h)^2+m

steel birch Feb 3, 2023, 2:08 AM

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so

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the vertex is (0, 0)

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it's at the origin

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the graph is x^2

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the vertex form for a parabola is

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y = 1/4p(x-h)^2 + k

sinful shard Feb 3, 2023, 2:10 AM

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so 1/4 is a formula

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for a

steel birch Feb 3, 2023, 2:10 AM

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yes

sinful shard Feb 3, 2023, 2:10 AM

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but

steel birch Feb 3, 2023, 2:11 AM

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also where does it say that 1/4 is p?

sinful shard Feb 3, 2023, 2:11 AM

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I edited it rq

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I’m a little slow but

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why is 1/4 a even though it says 1

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I never understood why 🥲

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WAIT NEVERMIND

steel birch Feb 3, 2023, 2:14 AM

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1 is just the base value

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the f(x) = 1(x-0)^2+0

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is just saying f(x) = x^2

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so yeah

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if 1 is the value in front of x

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then p is 1/4

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wait no i mean p=4 i think

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because if you solve for p using 1 = 1/4p you get 4

sinful shard Feb 3, 2023, 2:17 AM

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you’re confusing yourself

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you have to cancel the common factor of 1 for that

steel birch Feb 3, 2023, 2:18 AM

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i probably am lol

sinful shard Feb 3, 2023, 2:18 AM

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1
4 • 1

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yeah

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:D! thank you for the help fr

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I’m pretty new to it being like that as well

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so it kinda confused me on why it was like that

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but I think I get it now

steel birch Feb 3, 2023, 2:19 AM

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i don't think i've seen it like that

sinful shard Feb 3, 2023, 2:19 AM

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fr?

steel birch Feb 3, 2023, 2:20 AM

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not really

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it makes sense with the 1

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but i've been mostly exposed to

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the vertex form, slope form, and standard

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this was what i was talking about

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if it was something like y = 1(x-3)^2 + 2

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the p value would be 4

sinful shard Feb 3, 2023, 2:23 AM

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p is uh

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parabola right

steel birch Feb 3, 2023, 2:23 AM

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no

upper field Feb 3, 2023, 2:23 AM

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Check #5,
(x,y)
x is correct, y is not

steel birch Feb 3, 2023, 2:23 AM

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it's the distance from the vertex to the directrix or focus

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the vertex to the directrix and vertex to the focus are equidistant

sinful shard Feb 3, 2023, 2:24 AM

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I’ll be sure to check #5

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the distance from the vertex to the focus

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wouldn’t the focus be

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(0,1/4)

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and the

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directrix -1/4

steel birch Feb 3, 2023, 2:27 AM

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wait where are you getting these p values from?

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how did you get p = 1/4

sinful shard Feb 3, 2023, 2:28 AM

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oh shoot

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that was an typo

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yeah you’re hella confused

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find y=a(x-h)^2+m

upper field Feb 3, 2023, 2:30 AM

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@sinful shard Do you have a problem that asks you for the focus? Are you referring to #3 or something else?

steel birch Feb 3, 2023, 2:30 AM

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yeah i'm getting really confused on what you're asking for

sinful shard Feb 3, 2023, 2:30 AM

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#3

upper field Feb 3, 2023, 2:30 AM

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steel birch Feb 3, 2023, 2:32 AM

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sinful shard #3

are you just filling out the table?

sinful shard Feb 3, 2023, 2:33 AM

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I’m graphing and filling the table out

steel birch Feb 3, 2023, 2:33 AM

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if you are then you can just plug in values around 0 for x into f(x)=x^2 to figure out the coordinates

upper field Feb 3, 2023, 2:33 AM

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Yes, what Packers said is all you need to do. It doesn’t ask you for the directrix or focus

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You actually don’t need those to graph a parabola. You can use the vertex and one other coordinate as the bare minimum. In this case, the table wants you to use a total of 5 coordinate points. Use x = -2, -1, 0, 1, 2
It’ll give you easy points to graph

steel birch Feb 3, 2023, 2:35 AM

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yep

upper field Feb 3, 2023, 2:36 AM

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Plug in your inputs, x, and then write all the outputs, y

sinful shard Feb 3, 2023, 2:46 AM

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I got it now

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it was this easy

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4,1,0,1,4

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for my y

upper field Feb 3, 2023, 2:47 AM

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Yes, that is all correct, nice job! 😁

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Make sure you plot all your points and draw your parabola going through them

sinful shard Feb 3, 2023, 2:49 AM

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I sure will

#find y=a(x-h)^2+m