Question

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

Answer 1

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

Answer 2

I think it is y=-2x-3

Answer 3

Oops I meant xD y=-2x+1

Answer 4

is that for vertex form or intercept form? :)

Answer 5

Oops damn I put that in slope-intercept... Geez I suck at math xD, I don't think I can help you on vertex

Answer 6

hahah thanks for trying :)

Answer 7

use the quadratic formula and find the x-intercepts first

Answer 8

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

Answer 9

sure :) u knw the quadratic formula ?

Answer 10

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

Answer 11

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

Answer 12

y=x^2-2x-3

a = ?
b = ?
c = ?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

b = -2

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

Answer 17

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

Answer 18

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

Answer 19

a = 1
b = -2
c = -3

Answer 20

use above,
take ur time :)

Answer 21

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


\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)

Answer 22

i got 1 ± 4 .... is that close? :o

Answer 23

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


\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)


\(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\)


\(\large x = \frac{2 \pm \sqrt{16}}{2}\)

Answer 24

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

Answer 25

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


\(\large x = \frac{--2 \pm \sqrt{(-2)^2-4 \times 1\times (-3)}}{2(1)}\)


\(\large x = \frac{2 \pm \sqrt{4 + 12}}{2}\)


\(\large x = \frac{2 \pm \sqrt{16}}{2}\)


\(\large x = \frac{2 \pm 4}{2}\)

\(\large x = 1 \pm 2\)

Answer 26

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

Answer 27

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

\(x = 3, -1\)

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

vertec form is also easy,

\(h\) lies exactly at the middle of both the intercepts :-

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

Answer 33

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

Answer 34

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

Answer 35

so h=1?

Answer 36

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

Answer 37

yay :D

Answer 38

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

Answer 39

you still need to find \(k\)

Answer 40

:)

Answer 41

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

Answer 42

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

Answer 43

y=x^2-2x-3

put x = 1

Answer 44

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

Answer 45

-4?

Answer 46

yup ! thats ur \(k\)

Answer 47

y=(x−1)^2+4 would be the answer? :)

Answer 48

who ate the - sign before 4 ?

Answer 49

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

Answer 50

yes :)

Answer 51

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? :)

Answer 52

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

Answer 53

wats the standard form ?

Answer 54

y=ax^2+bx+c?

Answer 55

y = ax^2 + bx + c

this is the standard form right ?

Answer 56

oh yes... thats right !

Answer 57

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

add 5 both sides

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

Answer 58

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

Answer 59

expand right side,

Answer 60

would it be y=-2x+1

Answer 61

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

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

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

Answer 62

simplify :)

Answer 63

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

Answer 64

correct

Answer 65

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

Answer 66

yesss :D

Answer 67

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

\(\pm \sqrt{5/2} = x+1\)

\( -1 \pm \sqrt{5/2} = x\)

Answer 68

those are our x-intercepts

Answer 69

can u put t he intercept form ?

Answer 70

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

Answer 71

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