Question

can someone help me with quadratics please? you can make your own examples and guide me to the the answer.

Answer 1

can you be more specific?

Answer 2

ok well I'm working on quadratics functions and i need help solving them. I don't have any specific examples I just want to practice them.

Answer 3

can you post a problem ?

Answer 4

ok

Answer 5

y = -16x^2 + 96x
for instance this one : above

Answer 6

and what is the question?

Answer 7

kimberly tracks free fall of object she models a graph I don't have the graph but...

Answer 8

if this doesn't help I can try and find some other one ...

Answer 9

i have a perfect one 𝑔𝑔 − 8𝑔𝑔 − 20.

Answer 10

some background:

a quadratic is the name for a "second order polynomial", which means the highest exponent is 2
the simplest quadratic is
y= x^2

the most general form is
\[ a x^2 + b x +c \]
where a, b, and c are numbers.

If you graph a quadratic, you get a parabola

A parabola looks like a U or a \( \cap\)

Answer 11

yeah negative and positive ...

Answer 12

those were whats on the graph ...

Answer 13

what about factoring quad's though?

Answer 14

lets say I want to use this .. Factor 𝑦𝑦 + 11𝑦𝑦 + 24.

Answer 15

thats one y each not two

Answer 16

to factor quadratics, and see how to do it, it is better to start backwards
say you have
(x+a)(x+b)
can you multiply that out ?

Answer 17

/uh well I can multiply x and x by distributing right?

Answer 18

use the distributive property
to see it more clearly, temporarily rename (x+b) as B
(x+a)*(x+b) becomes (x+a)*B
now distribute the B
what do you get ?

Answer 19

uh well i see that you can do x^a'b?

Answer 20

imlost

Answer 21

B*(x+a)
now multiply B times each term inside the parens (this is called "distributing")

Answer 22

yes but I'm not sure whether its x^b + a^b or ...

Answer 23

theres no variable for B so idk what I'm multiplying.

Answer 24

the ^ means exponents. it is neither

try it with numbers
2*( 3+5)
multiply 2 times each number inside the parens

Answer 25

see now that theres variables it looks more reasonable that what i was understanding ... this is 6 + 10 = 16 after you've distributed.

Answer 26

/wasnt*

Answer 27

yes, and of course it is the same as 2* (3+5) = 2* 8= 16
so 16 both ways.
if you have letters, multiplying is easy... for example x times y is written xy
just like 2 times 3 is 2*3
(getting 6 is the next step, "simplifying" )
or 2 times x is written 2x

Answer 28

well I write that earlier I thought you disagreed with it...

Answer 29

wrote*

Answer 30

try
B*(x+a)

Answer 31

so this is bx + ab?

Answer 32

or xb*

Answer 33

yes. But I used B, because B is short for (x+a)

and bx= xb (order does not matter when multiplyiing: 2*3= 3*2)

Answer 34

so
B*(x+a)= Bx + Ba
and if we replace B with (x+b)
(x+b)(x+a)= (x+b)x +(x+b)a
or
(x+b)(x+a)= x(x+b) +a(x+b)

Answer 35

*B is short for (x+b) (not x+a )

Answer 36

/so much wisdom its making my brain fall out xD. thanks tough ... so what if you have me those letters mixed with integers also what do i do then?

Answer 37

gave*

Answer 38

though*

Answer 39

Unfortunately, you have to know this stuff before tackling quadratics.
you may not get it in one session, but you will get closer.

meanwhile
(x+b)(x+a)= x(x+b) +a(x+b)
now distribute x in the first x(x+b) and distribute the a in the second
can you do that ?

Answer 40

I can certainly try.

Answer 41

uh so i think its x^2 + b
and ax + b?

Answer 42

or am i missing something?

Answer 43

if the problem was 2(2+3) you did 2*2 + 3

Answer 44

yeah? thats 4 + 3? but Im not sure what I missed in the first one?

Answer 45

2(2+3) is 10
2*5=10
or distributing: 2*2 + 2*3 = 4+6 = 10

Answer 46

oh distribute both...

Answer 47

if we had x(x+b) we multiply x times *each* term inside the parens

Answer 48

so X^2 and xb?

Answer 49

yes

Answer 50

ok

Answer 51

(x+b)(x+a)= x(x+b) +a(x+b)
(x+b)(x+a)= x^2 +bx + ax + ab

Answer 52

so what about both numbers and letters can you perhaps explain those?

Answer 53

if we "factor out" an x from the 2 middle terms (the opposite of distribute)
we get
(x+b)(x+a)= x^2 +(b+a)x + ab

Answer 54

now we can do factoring of a quadratic
say we have
x^2 +7x +12

and we are trying to find a and b so (x+a)(x+b)= x^2 +7x +12

Answer 55

using the letters
x^2 +(b+a)x + ab

notice that (a+b) matches the 7 and a*b matches the 12 in
x^2 +7x +12

Answer 56

the first thing I do is list all pairs of numbers that when multiplied give us 12
1,12
2,6
3,4

Answer 57

then I add each pair
1+12= 13
2+6= 8
3+4= 7

if I find a pair that add up to the middle number (7 in this case) then we hit the jackpot
the factors are
(x+3)(x+4)
and if we multiply that out we will get x^2 +7x+12

Answer 58

alright... that was a bit of a long shot does it normally take that much steps?

Answer 59

it is like anything you learn. You probably forgot how hard it was to learn to read.
But in school, if you work at this every day, (and not try to learn it all in 1 day as you seem to be trying to do now) , it can be done.

Answer 60

try factoring x^2+4x+4

Answer 61

well uh ... I assume its 4^2x? for the first part

Answer 62

then add the 4?

Answer 63

factoring x^2+4x+4 means
find a and b so that
(x+a)(x+b)= x^2+4x+4

Answer 64

the first step is look at the last number: +4
find all pairs of numbers that multiply to give you 4
1,4
2,2

add each pair
1+4=5
2+2= 4

choose the pair that add up to the middle number (the 4 from 4x)
2,2 are the numbers for a and b
thus
(x+2)(x+2)= x^2 + 4x + 4

Answer 65

try factoring
x^2 +4x + 3

Answer 66

back

Answer 67

ok

Answer 68

wait idk how to factor this my teacher has asked me this before and i failed

Answer 69

x^2 +4x + 3

what is the first step? (look what I posted up above)

Answer 70

except she did it with a minus

Answer 71

i really don't know please walk me through ...

Answer 72

all i know is that I factor it ..

Answer 73

the first step is look at the last number:

Answer 74

3?

Answer 75

find all pairs of numbers that multiply to give you 3

Answer 76

3 and 1

Answer 77

now add them. if they add to the "middle number" 4, you found the correct pair

Answer 78

k

Answer 79

3+1= 4 (which matches the 4 in x^2 + 4x +3)
so 3 , 1 are the numbers we want
the factors are
(x+3)(x+1)

Answer 80

uh ok

Answer 81

try
y^2 + 11y + 24

which you posted at the start

Answer 82

do I start with 24

Answer 83

yes

Answer 84

in that case 12 and 2 can work..

Answer 85

or 24 ad 1

Answer 86

I always start with 1, 2, 3, etc
up to the square root of 21 (less than 5)

Answer 87

1,24
2,12
3,8
4,6

Answer 88

ok

Answer 89

so i pick 12 and 2

Answer 90

do any of the pairs add up to the middle number in y^2 +11y + 24

Answer 91

no

Answer 92

so is it done already

Answer 93

/thats been simplified?

Answer 94

not yet. recheck each pair

Answer 95

ok im lost Im not sure I see that that pairs except 12 and 2 equally the 24

Answer 96

add each pair
1,24
2,12
3,8
4,6

Answer 97

ill be back in two hours

Answer 98

1+24= 25
2+12= 14
3+8= 11
4+6= 10