Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Please help! how do you write y=x^2-2x-3 in vertex and intercept form using the quadratic formula?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

@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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

is that for vertex form or intercept form? :)
Oops damn I put that in slope-intercept... Geez I suck at math xD, I don't think I can help you on vertex
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
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
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 :)
\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)
i got 1 ± 4 .... is that close? :o
\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\) \(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\) \(\large x = \frac{2 \pm \sqrt{16}}{2}\)
thats close, just a minor mistake u did... check once
\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\) \(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\) \(\large x = \frac{2 \pm \sqrt{16}}{2}\) \(\large x = \frac{2 \pm 4}{2}\) \(\large x = 1 \pm 2\)
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 :(
vertec form is also easy, \(h\) lies exactly at the middle of both the intercepts :- \(h \) = \(\large \frac{3 + -1}{2}\)
\(h \) = \(\large \frac{3-1}{2}\)
\(h \) = \(\large \frac{2}{2}\)
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 = ?
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 :)
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? :)
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
wats the standard form ?
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?
standard form : y = -2x^2-4x+3
yesss :D
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 \(\pm \sqrt{5/2} = x+1\) \( -1 \pm \sqrt{5/2} = x\)
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question