ali1029
Please help! how do you write y=x^2-2x-3 in vertex and intercept form using the quadratic formula?
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ali1029
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@ganeshie8 @satellite73 do you guys know how to do this? :o
Mttblink
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I think it is y=-2x-3
Mttblink
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Oops I meant xD y=-2x+1
ali1029
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is that for vertex form or intercept form? :)
Mttblink
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Oops damn I put that in slope-intercept... Geez I suck at math xD, I don't think I can help you on vertex
ali1029
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hahah thanks for trying :)
ganeshie8
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use the quadratic formula and find the x-intercepts first
ali1029
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can you help me figure out how to do that? thats where im stuck :(
ganeshie8
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sure :) u knw the quadratic formula ?
ali1029
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sorta :o im just so confused because we've been using so many different forms!
ganeshie8
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dont wry, this is going to be very easy :)
First, we find x-intercepts using the quadratic formula.
ganeshie8
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y=x^2-2x-3
a = ?
b = ?
c = ?
ganeshie8
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\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
ganeshie8
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you need to figure out a, b, c values.
and plug them above to get the x-intercepts
ali1029
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so a=1 b=2 c=3? what i dont get is that it wants me to put that equation in intercept and vertex form but those have h & k :o
ganeshie8
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b = -2
we will get to vertex/intercept forms later.
first, get the x-intercepts :)
ganeshie8
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put them in the formula above,
wat do u get for x = ?
ali1029
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1 becaue 1^2 is 1? let me do that equation
ganeshie8
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a = 1
b = -2
c = -3
ganeshie8
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use above,
take ur time :)
ganeshie8
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\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)
ali1029
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i got 1 ± 4 .... is that close? :o
ganeshie8
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\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)
\(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\)
\(\large x = \frac{2 \pm \sqrt{16}}{2}\)
ganeshie8
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thats close, just a minor mistake u did...
check once
ganeshie8
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\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)
\(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\)
\(\large x = \frac{2 \pm \sqrt{16}}{2}\)
\(\large x = \frac{2 \pm 4}{2}\)
\(\large x = 1 \pm 2\)
ali1029
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ohhh i see what i did wrong! thank you :) so know what do i do?
ganeshie8
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that gives, \(x = 1+2 ~~~ or~~~~ x = 1-2\)
\(x = 3, -1\)
ganeshie8
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we're done wid using quadratic formula.
we got the x-intercepts : 3, -1
ganeshie8
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so the intercept form would be : y = (x-3)(x+1)
ganeshie8
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wat about vertex form ?
how to get \(h\), \(k\) ... any ideas ?
ali1029
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wow that was easier than i thought! thank you :) & no, not too sure :(
ganeshie8
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vertec form is also easy,
\(h\) lies exactly at the middle of both the intercepts :-
\(h \) = \(\large \frac{3 + -1}{2}\)
ganeshie8
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\(h \) = \(\large \frac{3-1}{2}\)
ganeshie8
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\(h \) = \(\large \frac{2}{2}\)
ali1029
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so h=1?
ganeshie8
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\(h \) = \(\large 1\)
ali1029
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yay :D
ganeshie8
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so vertex form wud be : \(y = (x-1)^2 + k\)
ganeshie8
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you still need to find \(k\)
ganeshie8
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:)
ali1029
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okay, so how do we find k now? :)
ganeshie8
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put x = h = 1, in given quadratic.
it gives u the \(k\) value
ganeshie8
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y=x^2-2x-3
put x = 1
ganeshie8
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y = 1^2 - 2(1) - 3
= 1 -2 -3
= ?
ali1029
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-4?
ganeshie8
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yup ! thats ur \(k\)
ali1029
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y=(x−1)^2+4 would be the answer? :)
ganeshie8
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who ate the - sign before 4 ?
ali1029
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i messed up again ._. so it would be y=(x−1)2-4?
ganeshie8
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yes :)
ali1029
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thank you so much!! you're amazing!! :) if it isnt too much too ask could you help me with two more equations similar to that? :)
ali1029
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i need to figure out how to put y-5=-2(x+1)^2 in standard and intercept form & y=(x+2)(x-3) in standard and vertex form :o
ganeshie8
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wats the standard form ?
ali1029
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y=ax^2+bx+c?
ganeshie8
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y = ax^2 + bx + c
this is the standard form right ?
ganeshie8
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oh yes... thats right !
ganeshie8
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y-5=-2(x+1)^2
add 5 both sides
y = -2(x+1)^2 + 5
ganeshie8
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next, use the formula \((a+b)^2 = a^2 +b^2 + 2ab\)
ganeshie8
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expand right side,
ali1029
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would it be y=-2x+1
ganeshie8
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y = -2(x+1)^2 + 5
= -2(x^2 + 1^2 + 2x ) + 5
= -2(x^2 + 1 + 2x) + 5
ganeshie8
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simplify :)
ali1029
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would it be -2x^2-4x+3?
ganeshie8
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correct
ganeshie8
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standard form : y = -2x^2-4x+3
ali1029
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yesss :D
ganeshie8
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y-5=-2(x+1)^2
to put this guy in intercept form, u need to find its x-intercepts.
so put y = 0
0-5=-2(x+1)^2
5/2 = (x+1)^2
\(\pm \sqrt{5/2} = x+1\)
\( -1 \pm \sqrt{5/2} = x\)
ganeshie8
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those are our x-intercepts
ganeshie8
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can u put t he intercept form ?
ali1029
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the intercept form is y=a(x-p)(x-q)
ali1029
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so if the equation is −1±5/2, how do i solve the fraction part?