# Graph this ?

1. Sep 7, 2004

### ACLerok

graph this plz?

How do I graph x=y-y^2? what would it look like? my calc is extremely rusty so if you can help me out, that'd be great. Thanks!

2. Sep 7, 2004

### Tide

It's a parabola with a vertex at (0.75, 0.5). It crosses the x axis at x = 0 and it crosses the y axis at y = 0 and y = 1. Can you take it from there?

3. Sep 8, 2004

### HallsofIvy

I don't get the vertex that Tide did.
The "base" parabola, of the form x=y2 (or y= x2) has vertex at (0,0) because if x is not 0, x2 is positive and not below 0. x=y-y2 is not a "perfect square" but you can complete the square: x= -(y2- y+ 1/4) +1/4 (half of the coefficient of y is -1/2 and the square of that is 1/4 so I add and subtract 1/4. When I take the -1/4 out of the parentheses, it is multiplied by that leading -1). The point of that is that -(y2- y+ 1/4)= -(y- 1/2)2, a perfect square.
We now have x= -(y- 1/2)2+ 1/4. If y= 1/2, then x= 1/4 (not 3/4). If y is any other number, (y-1/2)2 is positive so -(y-1/2)2 is negative and x is less than 1/4. The vertex is (1/4, 1/2). Of course, its easy to see, since x= y- y2= y(1- y), that if y= 0 or 1, x is 0. Knowing that the parabola passes through (0,1), (1/4,0), and (0,0) and x is never larger than 1/4 should make it easy to draw.

4. Sep 8, 2004

### Tide

Note to self: $\frac {1}{2} - \frac {1}{4} = \frac {1}{4}$

(Thanks, Halls!)

5. Sep 9, 2004

### ACLerok

how would i enter this in a graphics calculator to graph it?

6. Sep 9, 2004

### Tide

There are two ways to do that

(a) Interchange the labels x and y and enter them into your calculator recognizing that the x axis on your display is really the y axis and likewise for the y axis.

(b) Solve your original equation for y in terms of x and plot each of the two equations on the same graph.

7. Sep 9, 2004

### ACLerok

so if i use option b, i graph y=x and y=1-x ?

8. Sep 10, 2004

### Tide

No, you solve the quadratic equation $y^2-y + x = 0$ for y and plot the two y's that you get from that.