Given the following linear function, sketch the graph of the function and find the domain and range. f(x) = 2/3x - 3

1. To sketch the function f(x) = (2/3)x - 3, we first need to find two points that we can later join to sketch the line, for example the x- and y-intercepts.

a) The x-intercept occurs when f(x) = 0, so if f(x) = 0, then:

f(x) = (2/3)x - 3

0 = (2/3)x - 3

3 = (2/3)x (Add three to both sides)

3*(3/2) = x (Multiply both sides by 3/2)

9/2 = x

We have now found the x-intercept at (9/2, 0)

b) The y-intercept occurs when x = 0, so:

f(x) = (2/3)x - 3

f(0) = (2/3)*0 - 3

f(0) = -3

Now we know that the y-intercept is at (0, -3)

c) All that's left is to sketch the graph axes and label them, plot the two points, join them together using a ruler and label their coordinates.

2. The Domain is the range of x-values for which the function exists, and the Range is the range of y-values for which the function exists.

Since there haven't been any constraints specified, we can say that both the Domain and Range are (-∞, ∞), since the graph continues forever both along the x- and y-axis.

(Note that this isn't always the case and would change if, for example, the question stipulated that there was a domain of [0, 5] and you had to find the range. Then, you would calculate the value of y at each end of the domain (if x = 0, y = -3 and if x = 5, y = 1/3) - in my example, the range would thus be [-3, 1/3].)