Question

Rewrite parabola equations in standard form:
1. x^2+14x+44y+313=0
2. 3y^2+6y+108x-969=0
3. x^2-12x-48y-373=0

Answer 1

solve each for "y"

Answer 2

standard form => \(\bf y = ax^2\pm bx\pm c\)

Answer 3

I'm so confused... Can you please show me how to do the first one?

Answer 4

My teacher says I need to start by finding the vertex first...

Answer 5

There are many ways to find the equation.
You may use whichever you like.

Answer 6

Can you please help me and show me how to do the first one by finding the vertex?

Answer 7

hmm the "vertex form" and the "standard form" are no exactly the same.... -> http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms-thumb.png

maybe you're just asked to get the vertex form?

Answer 8

In order to find the standard form she says I must find the vertex and focus first, but answers are in fact supposed to be in standard form

Answer 9

or maybe he/she meant \(\Large \bf (x-h)^2=4p(y-k) \quad ?\)

Answer 10

ahhh the vertex and focus. point... then yes.... that will be \(\large \bf (x-h)^2=4p(y-k)\)

Answer 11

Yes thats it! How do I do it? haha

Answer 12

you know what a "perfect square trinomial" is, right?

Answer 13

I don't think so...

Answer 14

hmm lemme do a quick rundown on that... a perfect square trinomial
is a TRInomial
thus 3 terms

it comes from multiplying 2 EQUAL binomials

so \(\Large \bf (a\pm b)(a\pm b)\to (a\pm b)^2=a^2\pm 2ab+b^2\)

Answer 15

see the TRInomial?

Answer 16

Yes, that makes sense

Answer 17

so the trinomial rule is pretty simple if you look at the two terms in the binomial

so \(\Large \bf ({\color{blue}{ a}}\pm {\color{brown}{ b}})^2={\color{blue}{ a}}^2\pm 2\cdot {\color{blue}{ a}}{\color{brown}{ b}}+{\color{brown}{ b}}^2\)

notice the middle term
is really just 2 times both terms

and notice the left and right sides of the trinomial
is just each term squared

Answer 18

so with that in mind.... let us grab the 1st equation

Answer 19

Ok perfect

Answer 20

\(\bf x^2+14x+44y+313=0\implies x^2+14x=-44y-313
\\ \quad \\
(x^2+14x+{\color{red}{ \square ? }}^2)=-44y-313\)

any ideas what our missing fellow there is?

wel really need a number there to "complete the square"

so we need a number there, that multiplied times the 1st term and 2, will give us the middle term

Answer 21

49?

Answer 22

49? let's try that

Answer 23

Oh wait, so it would have to be a number squared... um maybe 7?

Answer 24

\(\large x^2+14x+{\color{red}{ 49 }}^2\implies
\begin{cases}
middle\ term\\
\hline\\
2\cdot x\cdot 7\implies 14x
\end{cases}\

yeap, is 49, or \(7^2\)

now, keep in mind that all we're doing is really "borrowing from our good friend, Mr Zero, 0"

so if we ADD \(\7^2\) we also have to SUBTRACT \(\7^2\)

so \(\bf x^2+14x+44y+313=0\implies x^2+14x=-44y-313
\\ \quad \\
(x^2+14x+{\color{red}{ 7 }}^2)-{\color{red}{ 7 }}^2=-44y-313
\\ \quad \\
(x+7)^2-49=-44y-313\implies (x+7)^2=-44y-313+49
\\ \quad \\
(x+7)^2=-44y-264\implies (x+7)^2=-44(y-6)\)

Answer 25

hmm anyhow .. tis a bit broken but \(\bf \large x^2+14x+{\color{red}{ 49 }}^2\implies
\begin{cases}
middle\ term\\
\hline\\
2\cdot x\cdot 7\implies 14x
\end{cases}\)

yeap, is 49, or \(7^2\)

so we add \(7^2\) and we subtract \(7^2\)

Answer 26

So, (x+7)^2=-44(y-6)
How would I get the final answer from here, or where does the 4 come in from (x-h)^2=4p(y-k)?

Answer 27

hmmm should be +6 actually since the negative is outside... so \(\bf x^2+14x+44y+313=0\implies x^2+14x=-44y-313
\\ \quad \\
(x^2+14x+{\color{red}{ 7 }}^2)-{\color{red}{ 7 }}^2=-44y-313
\\ \quad \\
(x+7)^2-49=-44y-313\implies (x+7)^2=-44y-313+49
\\ \quad \\
(x+7)^2=-44y-264\implies (x+7)^2=-44(y+6)\)

Answer 28

hmm ohh yeah... the 4... well the form is really \(\bf (x+7)^2=-44(y+6)\)

the "4p" just happen to be equal to -44

Answer 29

Ok! Thank you so much! It makes sense now, but I'm a little confused on the second one now with the 3y^2 and etc

Answer 30

the 2nd one is the same
notice the squared variable however is the "y" NOT the "x"

that just means is a "horizontal" parabola