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@ganeshie8 @satellite73 do you guys know how to do this? :o

I think it is y=-2x-3

Oops I meant xD y=-2x+1

is that for vertex form or intercept form? :)

hahah thanks for trying :)

use the quadratic formula and find the x-intercepts first

can you help me figure out how to do that? thats where im stuck :(

sure :) u knw the quadratic formula ?

sorta :o im just so confused because we've been using so many different forms!

dont wry, this is going to be very easy :)
First, we find x-intercepts using the quadratic formula.

y=x^2-2x-3
a = ?
b = ?
c = ?

\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

you need to figure out a, b, c values.
and plug them above to get the x-intercepts

b = -2
we will get to vertex/intercept forms later.
first, get the x-intercepts :)

put them in the formula above,
wat do u get for x = ?

1 becaue 1^2 is 1? let me do that equation

a = 1
b = -2
c = -3

use above,
take ur time :)

i got 1 ± 4 .... is that close? :o

thats close, just a minor mistake u did...
check once

ohhh i see what i did wrong! thank you :) so know what do i do?

that gives, \(x = 1+2 ~~~ or~~~~ x = 1-2\)
\(x = 3, -1\)

we're done wid using quadratic formula.
we got the x-intercepts : 3, -1

so the intercept form would be : y = (x-3)(x+1)

wat about vertex form ?
how to get \(h\), \(k\) ... any ideas ?

wow that was easier than i thought! thank you :) & no, not too sure :(

\(h \) = \(\large \frac{3-1}{2}\)

\(h \) = \(\large \frac{2}{2}\)

so h=1?

\(h \) = \(\large 1\)

yay :D

so vertex form wud be : \(y = (x-1)^2 + k\)

you still need to find \(k\)

:)

okay, so how do we find k now? :)

put x = h = 1, in given quadratic.
it gives u the \(k\) value

y=x^2-2x-3
put x = 1

y = 1^2 - 2(1) - 3
= 1 -2 -3
= ?

-4?

yup ! thats ur \(k\)

y=(x−1)^2+4 would be the answer? :)

who ate the - sign before 4 ?

i messed up again ._. so it would be y=(x−1)2-4?

yes :)

wats the standard form ?

y=ax^2+bx+c?

y = ax^2 + bx + c
this is the standard form right ?

oh yes... thats right !

y-5=-2(x+1)^2
add 5 both sides
y = -2(x+1)^2 + 5

next, use the formula \((a+b)^2 = a^2 +b^2 + 2ab\)

expand right side,

would it be y=-2x+1

y = -2(x+1)^2 + 5
= -2(x^2 + 1^2 + 2x ) + 5
= -2(x^2 + 1 + 2x) + 5

simplify :)

would it be -2x^2-4x+3?

correct

standard form : y = -2x^2-4x+3

yesss :D

those are our x-intercepts

can u put t he intercept form ?

the intercept form is y=a(x-p)(x-q)

so if the equation is −1±5/2, how do i solve the fraction part?