Question

Need help with quadratic function

Answer 1

for standard form i have 2(x+1)^2+1 and vertex is (1,-3/2) im not sure if its correct

Answer 2

You solved using completing the square right?

Answer 3

yes

Answer 4

Wouldn't the vertex just be \((h, k)\)?

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

So just \(\large (-1, 1) \)?

Answer 5

yah i was wondering that too

Answer 6

i did the vertex formula (-b/2(a), c -b^2/4(a)

Answer 7

so im kinda confused right now

Answer 8

oh never mind i did my vertex wrong with the fraction it is -1,1

Answer 9

Did you use the regular equation? (\(\large 2x^2+4x+3=0 \)) or the completing the square one?

Answer 10

i did y=2x^2+4x+3

Answer 11

and got 2(x+1)^2+1

Answer 12

Yea, they should turn out the same no matter what equation you use. I think the vertex form is easier tho :P

Answer 13

yeah i just redo my vertex formula and got -1,1 just now

Answer 14

thanks for your help

Answer 15

No problem, and welcome to OpenStudy :)

Answer 16

to find x/y intercept i have to substitute 0 w/ x right

Answer 17

thanks

Answer 18

Yes, to find the x-intercept set y=0
To find the y-intercept set x=0

:)

Answer 19

alright thanks

Answer 20

You're very welcome!