anonymous
  • anonymous
Please help! how do you write y=x^2-2x-3 in vertex and intercept form using the quadratic formula?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@ganeshie8 @satellite73 do you guys know how to do this? :o
anonymous
  • anonymous
I think it is y=-2x-3
anonymous
  • anonymous
Oops I meant xD y=-2x+1

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

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