Question

algebra 2 help

Answer 1

yes, I take two with whipped cream and chocolate chips sprinklees

Answer 2

\[x ^{2}+y^{2}+3x-4y-9=0\]

Answer 3

@jdoe0001 help? please :D

Answer 4

ok...so... hmmm what are you meant to do there?

Answer 5

identify the conic section and sketch the graph. if it is a parabola, give the vertex. If its a circle, give the center and radius. If its an ellipse or a hyperbola, give the center and foci.

Answer 6

i have the answer i just need to know how to do it.

Answer 7

alrite
so

let us check hmm hmm ok... hmm

you know how to identify them by just looking at the terms, right?

Answer 8

no but i know that its a circle XD

Answer 9

hehehe... you mean. because you looked at the answer on the back of the book =)

Answer 10

yes :/

Answer 11

well. that's hmmm anyway... one sec

Answer 12

ok.

Answer 13

anyway I might as well type it in

so.. to identify them first

if the squared variables, have the same coefficients, is a circle
if there's only one squared variable, is a parabola
if the squared variables, have the same sign, is an ellipse
if the squared variables, have different signs, is a hyperbola

Answer 14

got it. ok whats next?

Answer 15

so... we have two squared variables, x and y
and both have the same coefficient, 1
and the coefficients have the same sign, positive in this case

so, yes, is a circle

now... hmmm do you know what a "perfect square trinomial" is?

Answer 16

no i do not

Answer 17

im not good at math

Answer 18

well.. how about this form \(\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad

% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)

does that ring a bell?

notice, is a squared binomial, expanded to a trinomial

Answer 19

yes

Answer 20

well.. the trinomial expanded from the squared binomial, is so-called "perfect square trinomial"

so... keep an eye on the 2nd term
the 2nd term is 2 * the other two guys, without the exponent

so... now, lemme do some grouping

Answer 21

\(\bf x^2+y^2+3x-4y-9=0
\\ \quad \\ \quad \\
(x^2+3x)+(y^2-4y)-9=0
\\ \quad \\
(x^2+3x+{\color{red}{ \square }}^2)+(y^2-4y+{\color{red}{ \square }}^2)-9=0\)

so... notice
we group the "x"'s first
and then the "y"'s

now.. we seem to be missing a number there, to get a "perfect square trinomial"
so...any ideas on those two fellows?

Answer 22

why are the missing numbers squared?

Answer 23

well... notice the perfect square "trinomial" above
in order for us to get a perfect square trinomial, they have to be :)

Answer 24

so is 3x and-4y the 2ab and the -2ab?

Answer 25

yeap

Answer 26

oh ok I'm with you now

Answer 27

alrite...so....any ideas on the missing values?

Answer 28

3 squared ?

Answer 29

and -4 squared ?

Answer 30

well... let's see
we on the 1st group
we have 3x as the 2nd term

so... that means that 3x = 2 * x * \(\square\)
and you said that is 3
so
3x = 2 * x * 3 <--- is it?

Answer 31

yes?

Answer 32

we could also check the 2nd group
we have -4y as the 2nd term

so, that means
-4y = 2 * y * \(\square\)

you said is 4
so -4y = 2 * y * 4 <--- is it?

Answer 33

i think so.

Answer 34

well, if we use those values

3x = 2 * x * 3
3x \(\ne\) 6x

and

-4y = 2 * y * 4
-4y \(\ne\) 8y

Answer 35

well, on the 2nd term
nevermind the minus btw
we're only concerned with the coefficient

anyway though 4y \(\ne\)8y

Answer 36

shouldn't we divide by 2 instead of multiply?

Answer 37

heheh
see why I mentioned to keep an eye on the 2nd term of the trinomial? :)

that's where the other tems come from, usually

Answer 38

wait I'm confused again

Answer 39

do we multiply or divide?

Answer 40

the middle term in a perfect square trinomial
is 2 * the other terms on the side, without the exponent

in the groups we have here, for "x" and "y"
we're missing a term
but we do have the middle term of the trinomial
which we can use to get the missing term

Answer 41

b/2a?

Answer 42

b/2a yes

Answer 43

3/2x?

Answer 44

lemme do say... "y"
to make it simple
well, just use an equation :)
we know that 2 * x * something = 4y

so \(\bf 2\cdot y\cdot \square = 4y\implies \square =?\)

Answer 45

sorry

Answer 46

\[2\left[ \right]=4\]

Answer 47

anyhow... i have to dash
you may want to cover the perfect square trinomial firstly
or check your book on "completing the square" :)

Answer 48

ok. thank you for all your help