Question

Help please, needs to be put it into x = -3(y-v)^2 + h form

Answer 1

they y value of the vertex goes in as "v" in your formula
the x value goes in as h

the vertex is (3,-3)
what do you get ?

Answer 2

-3(y+2)^2+0 ?????

Answer 3

x = -3(y-v)^2 + h

that is usually written as
x= a(y-h)^2 + k
and we know (h,k) (i.e. (3,-3)

try again.

Answer 4

-3(y-3)^3-3

Answer 5

I confused you. because this parabola is sideways
the (h,k) is reversed.
x= a( y-k)^2 + h

Answer 6

in other words, we put the "y value" of (3,-3) in as k
and the "x value" of (3,-3) as "h"
in
\[ x= a(y-k)^2 + h \]

Answer 7

x=a(y+3)+3

Answer 8

yes, with an exponent
\[ x= a(y+3)^2 + 3 \]

now we use the other point they give us (0, -2)

put in 0 for x, and -2 for y. can you do that ?

Answer 9

you should get
\[ 0 = a(-2+3)^2 + 3\]
now we "solve for a"

Answer 10

-a(1)^2-3=0

Answer 11

@mathmale can you help me

Answer 12

In what specific way would you like for me to help you?

Answer 13

Well @phi left off on me solving for a, to get the equation to be in x = -3(y-v)^2 + h form

Answer 14

I agree that the parabola is a horizontal one; it opens to the left.

phi's formula for this parabola, \[x= a(y-k)^2 + h\]

Answer 15

is 100% on target. Point out where you began to feel lost.

Answer 16

I just wanted to confirm it was right, because phi was helping me and was guiding me then all of a sudden disappeared thats why

Answer 17

I do the same thing sometimes. Take it easy.

Type your equation here.

Answer 18

-a(1)^2-3=0

Answer 19

that is what I got for my equation

Answer 20

But you haven't yet found the value of the coefficient "a." How would you go about doing that?

Answer 21

As phi has shared with you, the general form of the equation for this parabola is\[x= a(y-h)^2 + k \]

Answer 22

Okay so to solve for A?

Answer 23

1. You are told that the vertex is (3,-3). That means h=3 and k=-3. Please substitute those values into phi's suggested equation. Your result?

Answer 24

Replace h in the equation with 3 and k with -3.

Answer 25

zarkam?

Answer 26

You were a bit unhappy when phi 'disappeared;' now it appears that you yourself have 'disappeared.'

Answer 27

Sorry, but it's my turn to 'disappear.' Never fear, I'll be back on OpenStudy later today, and will continue to help you then.

Answer 28

@welshfella help :/

Answer 29

I left you with some instructions. Mind following them?

1. You are told that the vertex is (3,-3). That means h=3 and k=-3. Please substitute those values into phi's suggested equation. Your result?

Answer 30

X=a(y-3)^2-3

Answer 31

you got to

****
-a(1)^2-3=0
****

as you know, using PEDMAS you do parens and exponents first
so you do (1)^2 which means 1*1
so you have

-a * 1 - 3 =0
-a times 1 is just -a

so you have
-a - 3 = 0

add +3 to both sides
-a -3+3 = 0 + 3

on the left, -3 + 3 is 0, and -a +0 is -a
on the right you get 3

so you have
-a = 3
now multiply both sides by -1
-1*-a = -1*3

simplify to get
a= -3

now use that in your equation
x= a(y+3)^2 + 3

to get

x= -3(y+3)^2 + 3

as a check, I used Geogebra to plot it. It looks like the parabola in your question, so the equation is good.