Graphing x=y-y^2 to Understanding and Visualizing the Equation

[ACLerok]

graph this please?

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!

 

[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?

 

[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.

 

[Tide]

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

(Thanks, Halls!)

 

[ACLerok]

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

 

[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.

 

[ACLerok]

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.

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

 

[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.