The exercise only calls for us to represent the function in two other ways. We will here also demonstrate a third representation.

Equation

We are told that the function has a slope of

\frac{2}{5}

and that it passes through the point

(12, - 5) .

Using this information, we can write our function as an equation in point-slope form.

y - y_{1} y - (- 5) = m (x - x_{1}) ⇓ = \frac{2}{5} (x - 12)

Let's simplify this equation by using that subtraction of a negative can be written as an addition.

y - (- 5) y + 5 = \frac{2}{5} (x - 12) ⇓ = \frac{2}{5} (x - 12)

Graph

Another way to represent this function is by graphing it on a coordinate plane. We have the point

(12, - 5)

and a slope of

\frac{2}{5} .

Recall that the slope is the change in

y

divided by the change in

x .

m = \frac{2}{5} ⇓ \frac{Δ y}{Δ x} = \frac{2}{5}

We will now first plot the point

(12, - 5) .

Then we will, as indicated by the slope, take a step of

5

units to the right and

2

units up to find a second point.

The points (12,-5) and (17,-3) and a line through them.

By drawing a line through the points we have our graph.

Table of Values

Yet another way to represent our function is by using a table of values. If one point and the slope is known we can make a rule for finding more points by using that the slope

\frac{2}{5}

can be written as

change

in

y

divided by

change

in

x .

\frac{Δ y}{Δ x} = \frac{2}{5}

A new point can be found by adding

5

to the

x -

coordinate and

2

to the

y -

coordinate of the known point.

x_{1} + Δ x = x_{2} \Rightarrow x_{1} + 5 = x_{2} y_{1} + Δ y = y_{2} \Rightarrow y_{1} + 2 = y_{2}

When a new point has been found, this second point can be used to find a third point, and so on. Let's create our table by starting at the given point

(12, - 5) .

We will use the rule twice, making a table of values with three ordered pairs.

$x + Δ x$	$y + Δ y$	$(x, y)$
		$(12, - 5)$
$12 + 5$	$- 5 + 2$	$(17, - 3)$
$17 + 5$	$- 3 + 2$	$(22, - 1)$

Exercise