Question

Using a directrix of y = -3 and a focus of (2, 1), what quadratic function is created?

Answer 1

@pooja195

Answer 2

@this_emo_chic1023 @triciaal

Answer 3

@KamiBug

Answer 4

@One098

Answer 5

@TuringTest

Answer 6

@iambatman @TuringTest @Preetha @iGreen

Answer 7

Jk :p

Answer 8

http://hotmath.com/hotmath_help/topics/finding-the-equation-of-a-parabola-given-focus-and-directrix.html

Answer 9

We have the formula:

\((x-a)^2 + b^2 - c^2 = 2(b - c)y\)

Where (a, b) is the focus, and 'c' is the directrix.

Here, the focus is (2, 1) and the directrix is -3.

So can you plug in a = 2, b = 1, and c = -3? @Brostep0s

Answer 10

\[(x-2)^2 + 1^2 +3^2 = 2(1+3)y\]

Answer 11

Yep, you got it!

Answer 12

I guess we can simplify this..

Let's simplify \((x-2)^2\) first.

\((x - 2)^2 \rightarrow x^2 - 2^2 \rightarrow x^2 + 4\)

Plug that back in:

\(x^2 + 4 + 1^2 + 3^2 = 2(1 + 3)y\)

Now let's simplify \(2(1 + 3)y\)

\(2( 1 + 3) y \rightarrow [(2 \times 1) + (2 \times 3)]y \rightarrow (2 + 6)y \rightarrow 8y\)

Plug that back in:

\(x^2 + 4 + 1^2 + 3^2 = 8y\)

Can you simplify the exponents and combine the like terms?

Answer 13

Um..what are your options..? Because I'm not sure which form they want it in..but I'm pretty sure it's standard form(\(ax + bx^2 + c\)).

Answer 14

\[ f(x) = \frac{ 1 }{ 8 }(x - 2)2 - 1\]

Answer 15

That's how they want it/

Answer 16

Oh..that's vertex form..

Answer 17

Okay, well can you simplify \(x^2 + 4 + 1^2 + 3^2 = 8y\)?

Answer 18

y=\[y=\frac{ x^2 }{ 8 } +\frac{ 7 }{ 4 }\]

Answer 19

................

Answer 20

It's wrong, isn't it?

Answer 21

\(1^2 \rightarrow 1 \times 1 \rightarrow 1\)
\(3^2 \rightarrow 3 \times 3 \rightarrow 9\)

Plug those back in:

\(x^2 + 4 + 1 + 9 = 8y\)

Now can you add 4 + 1 + 9 to simplify it completely?

Answer 22

14?

Answer 23

Yep..so we have:

\(x^2 + 14 = 8y\)

What you did was correct though..you divided 8 to both sides..giving us:

\(y = \dfrac{x^2}{8} + \dfrac{14}{8}\)

And \(\dfrac{14}{8}\) simplifies to \(\dfrac{7}{4}\)..which gives us:

\(y = \dfrac{x^2}{8} + \dfrac{7}{4}\)

To write this in function notation, we just switch \(y\) with \(f(x)\):

\(f(x) = \dfrac{x^2}{8} + \dfrac{7}{4}\)

Answer 24

THAT'S EXACTLY WHAT I PUT THE FIRST TIME!

Answer 25

I know that..I said you were correct..lol.

Answer 26

oh. Wow... If we were doing this in public, we would be having one of those awkward silences in which a homosexual baby was born. Congrats.

Answer 27

........................

Answer 28

Erm...well I think I'm doing something wrong..hold on.

Answer 29

NEVERMIND! You broke the silence lol

Answer 30

This is my last question for the 2nd part of my segment exam, so if we could move this along, that would be great lol

Answer 31

Okay, well we can write:

\(\sqrt{(x + 2)^2 + (y + 1)^2} = \sqrt{(y - 3)^2}\)
^^^^^^^^^^^^^^^^^^
This is the distance formula.

Where '2' is the x-value of our focus, and '1' is the y-value of our focus, and '-3' is our directrix.

Then we write:

\((x + 2)^2 + y^2 + 4y + 4 = y^2 - 6y + 9\)

The \(y^2\)'s cancel out:

\((x + 2)^2 + 4y + 4 = - 6y + 9\)

Subtract 4y to both sides:

\((x + 2^2) + 4 = -10y + 9\)

Subtract 4 to both sides:

\((x + 2)^2 = -10y + 5\)

Now we factor \(-10y + 5\) which gives us \(5(-2x + 1)\)

So we have:

\((x + 2)^2 = 5(-2x + 1)\)

I don't know what to do from here ;-;

Answer 32

@ganeshie8 @TuringTest

Answer 33

Idk, I'm just gonna guess. Thanks for the help

Answer 34

Wait..what are your options?

Answer 35

Too late, & I got it wrong, but I got a B+ on the exam.

Answer 36

Oh.. B+ is very good..congratz.

Answer 37

@cwrw238 @zepdrix @iambatman