Use the following conditions to create a sketch of the function by hand.
a. vertical asymptote at x=3
b. x-intercept of 1 and 5
c. f(2)=f(4)=f(6)=3
d. increasing on [0,3) and [5,infinity)
e. decreasing on (3,5]
f. f(x) is an odd function
I don't know how to graph this. It ends up looking like a triangle with a line down it.
I am unfamiliar with this type of graph since I have never graphed with only a vertical asymptote and not a horizontal.
Think of it as two curves to draw.
One starts at x just greater than 0, somewhere in negative y, passes through (1,0) and (2,3) and off to infinity at x=3.
The other comes down from infinity at x=3, passing through (4,3), and just touches the x-axis at a turning point (5,0) before heading off to infinity again through (6,3).
What I can't see from the given description is how far into negative y it goes near x=0. It definitely must "start" negative, since it increases on [0,3) and passes through (1,0), but we don't know.
And just when you thought it was safe to put down your pencil :-) -- it's an odd function, so you have to draw it all again upside down on negative x, since it has 180-degree rotational symmetry around the origin!
To graph a function with the given conditions, let's break down the information into different parts and then combine them to create the sketch.
a. Vertical asymptote at x=3: A vertical asymptote occurs when the function approaches infinity or negative infinity as x approaches a specific value. This means that the function cannot have a value at x=3.
b. X-intercepts of 1 and 5: An x-intercept occurs when the function passes through the x-axis. So, the graph will intersect the x-axis at x=1 and x=5.
c. f(2)=f(4)=f(6)=3: This condition tells us that the function passes through the points (2,3), (4,3), and (6,3). We can plot these points on our sketch.
d. Increasing on [0,3) and [5,infinity): This means that the function's values are increasing on the intervals [0,3) and [5, ∞). So, the graph should be rising from left to right on those intervals.
e. Decreasing on (3,5]: This means that the function's values are decreasing on the interval (3,5]. So, the graph should be falling from left to right on this interval.
f. f(x) is an odd function: An odd function exhibits symmetry about the origin. This means that if (x, y) is a point on the graph, then (-x, -y) is also a point on the graph. This implies that the graph should be symmetrical with respect to the origin.
Now let's combine all this information to create the sketch:
- Since we know that there is an x-intercept at x=1 and x=5, we can start by plotting these points on the x-axis.
- Next, we plot the points (2,3), (4,3), and (6,3) based on the condition f(2)=f(4)=f(6)=3.
- Since the function is increasing on [0,3) and [5,infinity), we can draw a line that goes upwards from x=0 to x=3 and a line rising from x=5 to infinity.
- Since the function is decreasing on (3,5], we draw a line that falls from x=3 to x=5.
- Finally, since the function is odd, we make sure that the graph is symmetrical with respect to the origin.
Putting all these steps together, you should end up with a sketch that resembles a "V" shape with the vertex at x=3 and the x-intercepts at x=1 and x=5. The graph should also pass through the points (2,3), (4,3), and (6,3), and exhibit symmetry with respect to the origin.