Question

what is the equation of the directrix of a parabola defined by the equation y=x^2-2x-3

Answer 1

hmm have you covered parabolas yet?

Answer 2

Yes but I don't fully understand it

Answer 3

hmm ok.... are you supposed to get the vertex.... by factoring, or completing the square or graphing?

Answer 4

well... you're needing the directrix.... thus why I mentioned the vertex

since once you find the vertex form of the parabola, the directrix is also found

Answer 5

Ok do I have to factor it out

Answer 6

do yo know what a perfect square, or "perfect square trinomial" is?

Answer 7

nope, you don't, but you could

Answer 8

I heard of perfect square

Answer 9

ok.... so \(\large \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?

Answer 10

Yes looks familiar

Answer 11

ok so let us "complete the square

so let's see, we have \(\bf y=x^2-2x-3\implies y+3=(x^2-2x+{\color{red}{ \square }}^2)\)

so...what do you think we're missing there, to gert a "perfect square trinomial" ?

Answer 12

Umm is it 2

Answer 13

2 ok
well, notice, the middle term of the perfect square trinomial
is 2 * the other two terms on its sides, without the exponent

so 2 * x * 2 = 2x?
^
middle term

so... guess not..... thus is not 2 t hen

Answer 14

Is the missing term suppose to equal 3

Answer 15

notice
the middle term is " 2x "
that means that \(\bf 2\cdot x\cdot {\color{red}{ \square }}=2x\)

if we use 3 for the missing one
2 * x * 3 \(\ne 2x\)

Answer 16

So its 1

Answer 17

well... let's see
2 * x * 1 = 2x \(\Large \checkmark\)

so now we know is 1

so keep in mind that
all we're doing is borrowing from our good fellow Mr Zero, 0
thus, if we add \(1^2\) we also have to subtract \(1^2\)
thus

Answer 18

\(\bf y=x^2-2x-3\implies y+3=(x^2-2x+{\color{red}{ 1 }}^2)-{\color{red}{ 1 }}^2
\\ \quad \\
y+3(x-1)^2-1\implies y+4=(x-1)^2\implies 1(y+{\color{blue}{ (-4)}})=(x-{\color{brown}{ 1}})^2
\\ \quad \\

\begin{array}{llll}

4{\color{purple}{ p}}(y-{\color{blue}{ k}})=(x-{\color{brown}{ h}})^2\\
\end{array}

\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\
{\color{purple}{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\)

Answer 19

After I plug in the values how do I solve it

Answer 20

so our vertex is at ( 1, -4)

the template for the vertex form, uses ( y -h)
thus (y+4) is really ( y - (-4)) = (y+4)

and notice the coefficient in front of it , is 1
so

we know that "4p" is equals to 1
4p =1
p = 1/4

so

the vertex is at (1, -4)
and the directrix is 1/4 units from there

the parabola has a leading term coefficient that is positive, that means the parabola is going upwards

so that means the directrix is below the vertex