How do i graph this function?

Question

1) h(x)=2/x/, where the domain is (-1, 0, 1, 2)
2) f(x)=x-1, where the domain is (x/x is a real number)

Answer 1

To graph functions, first get a piece of graph paper or you can free hand the X-Y axis and plot the points between -5 and 5 on both axis.

1) h(x) = 2/x domain of x (-1, 0, 1, 2)

Find the coordinate pairs for each member of the domain.
h(-1) = 2/-1 --> (-1, -2)
h(0) = 2/0 --> undefined so no point on the graph
h(1) = 2/1 --> (1,2)
h(2) = 2/2 --> (2,1)

The graph will be the three points above. It is not a line since the domain is limited to the integers listed.


2) f(x) = x-1 domain is x/x is a real numbers (which excludes x=0 from the domain, but includes all other real numbers)

Since the equation is in slope-intercept form, i.e. y = mx + b
the slope of the line is m=1 and the intercept on the y-axis is b=-1

You can draw the line that way, or determine two points on the line like this:
f(-1) = -1 - 1 --> (-1,-2)
f(1) = 1 - 1 --> (1, 0)

Since the domain is the real numbers except 0, a solid line is drawn to extend beyond both points. Either way you draw the line, make an open point (ie. a circle) on the line where x=0. Don't draw the line through the open point. The line is terminated with arrows to indicate that it extend to infinity in both directions

Answer 2

1) h(x) = 2|x|
Either use
a) substitution: x = -1,0,1,2 then y = 2,0,2,4 respectively or
b) sketching: Sketch y=2x and then reflect this graph about the x-axis such that all of the graph is above the x-axis.

2) f(x) = x-1
Similarly, you can use substitution and plot out the line or you can use the fact that y=mx+c and m is the gradient while c is the y-intercept.

This line cuts the y-axis at -1 and cuts the x-axis at -1 too.

Answer 3

I'm not to sure of the second one because if it's x/x=1 then f(x) is just zero.

I think the second one starts from (-1,2) then goes all the way up to infinity at 0 then comes down to (1,2) from infinity and ends at (2,1).