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.

We can use the point slope form 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 use the GeoGebra figure.

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 $\Delta y = f(1)-f(-1) = 4-2=2.$

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 at a focus point between the two axes, consistent with $=m=-1$ and that the function is decreasing at a rate of one unit on the target value for every unit increase in the domain value.

An arrow through $x=0$ and the focus point finds $f(0) = 3$.

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 GeoGebra figure.

An arrow through the focus point with its head at $0$ on the target comes from $x=3$ on the domain axis.