Make sure to use both positive and negative values for x.
Example Table:
x
(0.5x-3)^3
y=(0.5x-3)^3
- 2
(0.5( - 2)-3)^3
- 64
0
(0.5( 0)-3)^3
- 27
2
(0.5( 2)-3)^3
- 8
4
(0.5( 4)-3)^3
- 1
6
(0.5( 6)-3)^3
0
10
(0.5( 10)-3)^3
8
12
(0.5( 12)-3)^3
27
Graph:
Description: See solution.
Practice makes perfect
Let's make a table of values to graph the given function. When you are making a table of values, make sure to use a variety of points, including negative and positive values.
x
(0.5x-3)^3
y=(0.5x-3)^3
- 2
(0.5( - 2)-3)^3
- 64
0
(0.5( 0)-3)^3
- 27
2
(0.5( 2)-3)^3
- 8
4
(0.5( 4)-3)^3
- 1
6
(0.5( 6)-3)^3
0
10
(0.5( 10)-3)^3
8
12
(0.5( 12)-3)^3
27
We will now plot the obtained points and connect them with a smooth curve. Consider also that this is an odd-degreepolynomial with a positive leading coefficient. This tells us about the end behavior of the function.
&f(x) → - ∞ as x → - ∞
&f(x) → + ∞ as x → + ∞
Let's draw the function.
Graph description
Notice that both the domain and the range of the given function are all real numbers.
Domain:& All real numbers
Range:& All real numbers
Recall that odd-degree functions will always have an odd number of real zeros. We see above that y=(0.5x-3)^3 intersects the x-axis at one point. By looking at the table we can state the values for the x- and y-intercept.
y-intercept:& y=- 27
x-intercept:& x=6
We can see that the given function increases on the interval (- ∞, ∞), and that it has no axis of symmetry.
