# Make y a funciton of x

1. Apr 18, 2005

### shoopa

same with y=4x-x^2 goes to x=

thank you for help!!

2. Apr 18, 2005

### Moo Of Doom

Always by completing the square for quadratic functions.

3. Apr 18, 2005

### snoble

The method is pretty straight forward. Just divide both sides so that x^2 has a coefficient of 1. Then add on a constant C to both sides so that the right hand side is a square (remember square quadratics have the form X^2 + 2*b + b^2 = (x+b)^2). Then you just need to square root both sides and then it becomes clear how to get the x alone. You may want to add a $$\pm$$ when you take the square root if you are doing that in this class.

so for y = ax^2 +bx just divide both sides by a, add (b/(2a))^2. Then the right hand side can be written out as (x + b/(2a))^2. Then just square root and add stuff around. Remember nothing says a or b can't be negative. a just can't be 0.

4. Apr 18, 2005

### dextercioby

You can express "x" in terms of "y" only on a certain interval.That is to say that the function "y(x)" is invertible only on an interval,and not on R.

Daniel.

5. Apr 18, 2005

### Berislav

Yes, but I think that that won't be a good function, since it doesn't map -to one on any interval.

EDIT:
Ah, I see now that you mean interval of x. Sorry.

Last edited: Apr 18, 2005
6. Apr 18, 2005

### eNathan

I think that I can correctly derive (without testing) that

$$x = \sqrt { \frac {y} {2} } + \frac {8} {y}$$

Last edited: Apr 18, 2005
7. Apr 18, 2005

### snoble

I don't understand. 8*1-2*1^2=6 but 6*1-1^2=5
So the two functions differ at 1 right?
The first case you should get $$x=2\pm 1/2\,\sqrt {16-2\,y}$$
and the second cased
$$x=3/4\pm 1/4\,\sqrt {9-4\,y}$$
Two different answers for two different equations.

8. Apr 19, 2005

### baby_garfield

y=8x-2x^2--> 8x-2x^2-y=0--> 2x^2-8x+y=0

9. Apr 19, 2005

### baby_garfield

never mind! that was wrong

10. Apr 19, 2005

### baby_garfield

no! it's not! it's right! (sorry for not being sure )

11. Apr 19, 2005

### dextercioby

'Twas not.It was correct.How else could this problem b solved?

Daniel.

12. Apr 22, 2005

### abia ubong

u should be strongwilled baby garfield, the step is right, the use of the quadratic formula gives x in terms of y and that satisfies ur problem dexter

13. Apr 22, 2005

### The Bob

$$y = 8x - 2x^2$$

$$\Rightarrow 2x^2 - 8x + y = 0$$

Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$

a = 2, b = -8 and c = y

$$x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}$$

$$\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}$$

$$\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}$$

Last edited: Apr 22, 2005
14. Apr 22, 2005

### The Bob

$$y = 4x - x^2$$

$$\Rightarrow x^2 - 4x + y = 0$$

Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$

a = 1, b = -4 and c = y

$$x = \frac{ 4 \pm \sqrt{(-4^2) - (4 \times 1 \times y)}}{2 \times 1}$$

$$\Rightarrow x = \frac{ 8 \pm \sqrt{16 - 4y}}{2}$$

$$\Rightarrow x = 4 \pm \sqrt{4 - y}$$