Question

(vertex form of a parabola) Write equations in vertex form with the given information.

a. vertex of (7,9), a=1

b. vertical stretch of 2, vertex of (-1,-3)

c. reflected across the x axis, vertex of (5,0)

Answer 1

so vertex form is

y=a(x-h)^2+k

a is the vertical stretch factor
if a is negative it gets reflected across the x axis

(h,k) is the vertex

can you try plugging some things in?

Answer 2

so it would be \[1(x-7)^{2}2+9\] ???

Answer 3

was the extra 2 a typo? and since its just a 1 you can just not even write it

y=(x-7)^2+9

Answer 4

What do you mean extra 2 doesnt ^ mean squared?

Answer 5

Yes, but you have another 2 after the parentheses

Answer 6

well you wrote y=a(x-h)^2+k is there not suppose to be a 2 at all?

Answer 7

yes there is but just an exponent

this is vertex form

\[y=a(x-h)^{2}+k\]

plugging things in from number 1 should give you

\[y=(x-7)^{2}+9\]

not

\[y=(x-7)^{2}2+9\]

Answer 8

ok thats the answer to a i understand that but what about b and c?

Answer 9

Try it out, what do you think b is?

Answer 10

do we do the same as the first one? so instead of 1 plug in 2? so 2(x+1)^-3 ??

Answer 11

really close

\[y=2(x+1)^{2}-3\]

Answer 12

isnt that what i wrote (not to sound rude) and then do you solve it from there or leave it as it is?

Answer 13

Question asks for vertex form and its in that form.

you had the -3 as the exponent, 2 is the exponent

remember vertex form is

\[ y=a(x-h)^{2}+k\]

k is the y coordinate of the vertex and it isnt in the exponent

Answer 14

Oh i understand what i did wrong and do i just leave the answer as it is or do i try to go further

Answer 15

Nope dont have to go any further.

so now try c

Answer 16

ok so (5,0) would it be (0,-5)

Answer 17

I have no idea what you just did =p

do it just like the others, except to reflect it across the x axis you make 'a' negative

Answer 18

Since theres no vertical stretch factor just slap a negative sign where a is

Answer 19

ok so it is\[-(x-5)^{2}+0\]

Answer 20

Yep and since its just a 0 you can pretend its not even there.

\[y=-(x-5)^{2}\]

Answer 21

Alright thank u so much

Answer 22

You're welcome :)