*Analytic equilibria.* Equilibria are values of $q_t$ where $f(q_t)=q_t$, i.e., find values of $q_t$ that when plugged into the function $f(x)=x^2-x$, we get $q_t$ back out again. The values $q_t$ must satisfy the equation
$$q_t^2 - q_t = q_t.$$
To stress that we are talking about equilibria, let's substitute $E$ for $q_t$, so the equation becomes

One thing we **cannot** do is divide both sides of the equation by $E$. Why can't we do that? We have to remember that we cannot divide both sides of the equation by

and it is possible that $E$ might be equal to

.

Instead, we need to set the equation equal to zero by subtracting

from both sides of the equation. The resulting equation is

$=0$.
Factor out the $E$ from the left hand size, obtaining:

$=0$
Since we have the product of two factors equal to zero, one of them must be zero, so we can conclude that

$=0$ or

$=0$.
Therefore, the equilibria are
$E=$

(separate answers by commas).
*Graphical equilibria.* A second way to find equilibria is via a plot of $q_{t+1}$ versus $q_t$ (i.e., have $q_t$ on the $x$-axis and $q_{t+1}$ on the $y$-axis). On the below axes, graph the function $q_{t+1}=f(q_t)$. (Recall that the function is $f(x)=x^2-x$.) Using a different color (or different line thickness), also plot the diagonal line $q_{t+1}=q_t$. (Online, drag the blue points so the thick blue curve is the graph of $q_{t+1}=f(q_t)$ and drag the green points so the thin green line is the graph of $q_{t+1}=q_t$.)
**Feedback from applet**

function:

intersections:

line:

Find the points where the diagonal and the graph of $f$ intersect and draw them on the graph. (Online, change the $n_i$ slider to indicate the number of intersections and drag the points that appear to the proper locations.) What are the coordinates of the points? If we label the points $A$ and $B$, the points are:

$A =$

and $B=$

At those points $(q_{t},q_{t+1})$, both $q_{t+1}=f(q_t)$ and $q_{t+1}=q_t$. This means that $q_{t}=f(q_t)$ (i.e., $f$ didn't change the value $q_t$). The value of $q_t$ at those points are the equilibria $E$ for our dynamical system. The equilibria are

$E=$

.