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can you be more specific?
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.
can you post a problem ?
y = -16x^2 + 96x for instance this one : above
and what is the question?
kimberly tracks free fall of object she models a graph I don't have the graph but...
if this doesn't help I can try and find some other one ...
i have a perfect one 𝑔𝑔 − 8𝑔𝑔 − 20.
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 \( \cap\)
yeah negative and positive ...
those were whats on the graph ...
what about factoring quad's though?
lets say I want to use this .. Factor 𝑦𝑦 + 11𝑦𝑦 + 24.
thats one y each not two
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 ?
/uh well I can multiply x and x by distributing right?
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 ?
uh well i see that you can do x^a'b?
B*(x+a) now multiply B times each term inside the parens (this is called "distributing")
yes but I'm not sure whether its x^b + a^b or ...
theres no variable for B so idk what I'm multiplying.
the ^ means exponents. it is neither try it with numbers 2*( 3+5) multiply 2 times each number inside the parens
see now that theres variables it looks more reasonable that what i was understanding ... this is 6 + 10 = 16 after you've distributed.
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
well I write that earlier I thought you disagreed with it...
so this is bx + ab?
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)
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)
*B is short for (x+b) (not x+a )
/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?
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 ?
I can certainly try.
uh so i think its x^2 + b and ax + b?
or am i missing something?
if the problem was 2(2+3) you did 2*2 + 3
yeah? thats 4 + 3? but Im not sure what I missed in the first one?
2(2+3) is 10 2*5=10 or distributing: 2*2 + 2*3 = 4+6 = 10
oh distribute both...
if we had x(x+b) we multiply x times *each* term inside the parens
so X^2 and xb?
(x+b)(x+a)= x(x+b) +a(x+b) (x+b)(x+a)= x^2 +bx + ax + ab
so what about both numbers and letters can you perhaps explain those?
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
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
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
the first thing I do is list all pairs of numbers that when multiplied give us 12 1,12 2,6 3,4
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
alright... that was a bit of a long shot does it normally take that much steps?
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.
try factoring x^2+4x+4
well uh ... I assume its 4^2x? for the first part
then add the 4?
factoring x^2+4x+4 means find a and b so that (x+a)(x+b)= x^2+4x+4
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
try factoring x^2 +4x + 3
wait idk how to factor this my teacher has asked me this before and i failed
x^2 +4x + 3 what is the first step? (look what I posted up above)
except she did it with a minus
i really don't know please walk me through ...
all i know is that I factor it ..
the first step is look at the last number:
find all pairs of numbers that multiply to give you 3
3 and 1
now add them. if they add to the "middle number" 4, you found the correct pair
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)
try y^2 + 11y + 24 which you posted at the start
do I start with 24
in that case 12 and 2 can work..
or 24 ad 1
I always start with 1, 2, 3, etc up to the square root of 21 (less than 5)
1,24 2,12 3,8 4,6
so i pick 12 and 2
do any of the pairs add up to the middle number in y^2 +11y + 24
so is it done already
/thats been simplified?
not yet. recheck each pair
ok im lost Im not sure I see that that pairs except 12 and 2 equally the 24
add each pair 1,24 2,12 3,8 4,6
ill be back in two hours
1+24= 25 2+12= 14 3+8= 11 4+6= 10