Question

Help @jim_thompson5910 @rajee_sam

Answer 1

this one is up and down i dont know it

Answer 2

the parabola can be rewritten in this equation: \[\large y=a(x-h)^2+k\] with your vertex at \[\large v=(h,k)\]

Answer 3

$$
x=a(y-\color{blue}{k})^2+\color{red}{h}\ \text{ opens to the right}\\
x=-a(y-\color{blue}{k})^2+\color{red}{h} \ \text{ opens to the left}\\
\text{vertex is at }(\color{blue}{k},\color{red}{h})
$$

Answer 4

woops, darn, wrong form lol

Answer 5

You the points of your vertex, substitute them into the equation and find your answer

Answer 6

$$
y=a(x-\color{blue}{h})^2+\color{red}{k}\ \text{ opens upward}\\
x=-a(x-\color{blue}{h})^2+\color{red}{k} \ \text{ opens downward}\\
\text{vertex is at }(\color{blue}{h},\color{red}{k})
$$

Answer 7

so much for pasting hehe

$$
y=a(x-\color{blue}{h})^2+\color{red}{k}\ \text{ opens upward}\\
y=-a(x-\color{blue}{h})^2+\color{red}{k} \ \text{ opens downward}\\
\text{vertex is at }(\color{blue}{h},\color{red}{k})
$$

Answer 8

how do you paste?

Answer 9

wait which one do i use?

Answer 10

theone that opens downward

Answer 11

i mean which one of the equations theres alot up there

Answer 12

@Jhannybean ctrl-v :/, I don't think is what you mean though

Answer 13

or you could use the equation i gave you,plug your points in,and find the equation you're looking for.
Jdoe's equations tell you how to find a vertex of a parabola that is facing upwards or downwards.

Answer 14

ok but there is 2 different ones he did there one that has x and y and the othere has y and y which one is correct to use with these type of questions

Answer 15

oh, you'd use the one for solving with y. by that i mean "y=" the last post he made.

Answer 16

the other one was incorrect right?

Answer 17

Yes.

Answer 18

IF you were to use the ones solving for "x", the parabola would be opening up left and right, NOT up and down. I believe.

Answer 19

ok thank you much understand able so i just have to plug it in?

Answer 20

$$
(\color{red}{h},\color{blue}{k}) \implies (\color{red}{x},\color{blue}{y}) \implies (\color{red}{-5},\color{blue}{4})\\
y=-a(x-\color{red}{h})^2+\color{blue}{k}\\
x=-a(x-\color{red}{(-5)})^2+\color{blue}{(4)}
$$
as @Jhannybean suggested

so, what do you get?

Answer 21

yes, just plug them in.

Answer 22

i got d?

Answer 23

yes :D

Answer 24

Given the standard equation for a parabola opening left or right, which way does a parabola open when the coefficient of the y2-term, a, is positive? Left or right?

this to?

Answer 25

$$
x=a(y-\color{blue}{k})^2+\color{red}{h}\ \text{ opens to the right}\\
x=-a(y-\color{blue}{k})^2+\color{red}{h} \ \text{ opens to the left}\\
\text{vertex is at }(\color{blue}{k},\color{red}{h})
$$

Answer 26

you tell me

Answer 27

well, vertex is ... wrong there, hold one hehe

Answer 28

i thing its left

Answer 29

$$
x=a(y-\color{blue}{k})^2+\color{red}{h}\ \text{ opens to the right}\\
x=-a(y-\color{blue}{k})^2+\color{red}{h} \ \text{ opens to the left}\\
\text{vertex is at }(\color{red}{h},\color{blue}{k})
$$

Answer 30

when a is positive, the parabola opens to the right, that means it takes on all the positive x values

when a is negative, it opens left, and takes on ALL the negative x-values.