Example LF.FORM.3 : Suppose $f$ is a linear function with
$f(1) = 2$ and $f(-1) = 4$. Find the slope and then
point-slope form of the line. Visualize $f$ with a mapping diagram
that illustrates the slope = $m$ with $f(-1) = 2$ and $f(1) =4$.
Find the X- and Y-intercepts for $f$ and visualize them on the
mapping diagram as well.
Explain how to find the focus point for $f$ and how this can be used
to find the intercepts and slope without using any of the forms or
equation to find slope.
Solution: The slope is the rate and magnification factor. $m
= \frac {\Delta y} {\Delta x} = \frac {f(1)-f(-1)} {1 - (-1)} =
\frac {2-4} 2 = -1$ .
We can use point slope with $m=1$ and $f(1)=2$ to obtain $f(x) = 2 -
1*(x-1) = - x + 3.$ This is confirmed by checking $f(-1) = 2 -
(-1-1) = 4$.
You can now work the remainder of the algebra part of problem as
with the other examples.
Draw a mapping diagram yourself or have SAGE create this by
evaluating. First use the two data points to determine the slope.
Given a point / number, $x$, on the source line, there is a unique
arrow meeting the target line at the point / number, $2 -
1*(x-1) = -x + 3$, which corresponds to the linear function's value
for $x$
When the point in the domain is $-1$, the arrow points to $f(-1) =
4$ visualizing the point $(-1,4)$ on the graph of $f$. We also have
$f(1)= 2$ so the difference in the value of $f$ for -2 units change
in $x$ is $f(1)-f(-1) = 4-2=2.$
Since a unit step is used in this mapping diagram, we see the slope
(magnification, rate) visualized in the gap between the heads of
consecutive arrows on the mapping diagram.
As $m < 0$, the arrows cross illustrating that the function
is decreasing at a rate of one unit on the target value for
every unit increase in the domain value.
The X- intercept is the value $a$ on the domain axis from which the
arrow hits the number $0$ on the target. For this function,
that value is $a= 3$. You can check this by entering $3$ for $a$ on
the SAGE results.
Since the two arrows from the domain points $x= -1$ and $x=1$ are
crossing,$f$ has the focus point consistent a slope $=m=-1$.
An arrow through $x=0$ finds $f(0) = 3$.
An arrow with its head at $0$ on the target can been seen to come
from $x=3$ on the domain axis.