Question

Find a formula for the function f that is defined by
y= f(x) <-> 2x+ 2xy+ y^2=5 and y> -x

Answer 1

What does <-> mean?

Answer 2

What are you looking for when you say "formula?" What mathematics course are you taking. That will help narrow it down.

Answer 3

<-> = if and only if @Loser66

Answer 4

I think it just means y= f(x) equals that equation. I just have to write a formula using the notation f(x) =something. Calculus and Analytical Geometry, we just work with functions that have one variable

Answer 5

So i think I need to get it in terms of y

Answer 6

Should I use the quadratic formula?

Answer 7

Thats what i was thinking

Answer 8

Think of the left side as a quadratic.

Answer 9

Yes.

Answer 10

So I have to get it in the form ax^2+bx+c=0. Would I divide the entire equation by 5?

Answer 11

yes move over the 5
so you would have
y^2+2xy+(2x-5)=0
where 2x-5=C

Answer 12

Nahhh you would subtract 5 from both sides
5-5=0

Answer 13

Think of "2x" as a constant. Factor the left side accordingly.

Answer 14

So would 2x be like the coefficient? when I'm trying to figure out what a, b, and c are to solve for quadratic formula

Answer 15

a=1 b=2x and c=2x-5

Answer 16

Oh. Thank you. So how am I supposed to give medals to you guys. What do I click?

Answer 17

Appreciate the help. All of you

Answer 18

No problem. Did the quadratic formula work out yet?

Answer 19

I think I got it from here. The rest is easier algebra

Answer 20

\[x= -x \pm(\sqrt{4x^2-8x+20}/2)\]

Answer 21

Okie doke. Not sure about the medals. I'm new here. Happy mathing!

Answer 22

\[-2x +\sqrt{-6x-20} / 2\]

Answer 23

Make sure you divide both the 2x and the sqrt by two.

Answer 24

\[\sqrt{b^2-4(a)(c)} = \sqrt{(2x)^2-4(1)(2x-5)} = \sqrt{4x^2 - 8x - 20}\]

Answer 25

My final answer: \[x \pm \sqrt{x^2-2x+5}\] as your two factors. You would have to multiply them to get your equation for f(x).

Answer 26

So.... \[(x +\sqrt{x^2-2x+5})(x-\sqrt{x^2-2x+5})\]