
A set of college algebra problems on graphs of functions with answers, are presented. The solutions are at the bottom of the page.

The graph of f(x) is shown below. Draw the graph of y = f(x  3)  2
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Complete the graph given below so that it is symmetric with respect to the origin.
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The graph of f(x) is shown below. Sketch the graph of y = f(x + 1)  1
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The graph of f(x) is shown below. Sketch the graph of the inverse of f.
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The graph of h(x) is shown below.
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a) evaluate: h(2) + h(2) =
b) What is the domain of h?
c) What is the range of h?
d) Find the interval(s) over which h is increasing.
e) Find the interval(s) over which h is decreasing.
f) Find the interval(s) over which h is constant.

The graph of f(x) is shown below. Sketch the graph of f(2x).
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The graph of f^{ 1}(x) is shown below.
Evaluate the following: f(0) , f(2)
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Answers to the Above Questions

Select points whose coordinates are easy to determine on the given graph (see graph in black below) and then transform them as follows:
1  shift right 3 units : f(x  3)
2 reflect on the xaxis :  f(x  3)
3  shift down 2 units :  f(x  3)  2
4  connect the transformed points to sketch the transformed graph shown in red below.
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A graph is symmetric with respect to the origin if for each point (a,b) on the graph there exists a point (a,b) on the same graph.
We select points (a,b) on the given graph and then transform them into (a,b) to obtain more points. When put together the graph is symmeteric with respect to the origin.(see the whole graph black and red below).
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Select points on the given graph and then transform them as follows:
1  reflect on the y  axis : f(x)
2 shift right one unit: f( x + 1) = f((x  1))
3  shift down 1 unit : f( x + 1)  1
4  connect the transformed points to sketch the transformed graph shown in red below.
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We first determine points (a , b) on the graph of the given function and then use the definition of the inverse to determine points (b , a) on the graph of the inverse. Or use the line y = x to reflect points (a , b) into (b , a).
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 h(2) + h(2) = 3 + 1 = 2
 Domain: [5 , 5]
 Range: {3} U [0 , 5]
 increase: [3 , 2) , [2 , 5]
 decrease: [5 , 3]
 constant: [2 , 2)

Function f has two xintercepts: x = 2 and x = 2 . f(2x) will also have xintercepts such that 2x = 2 which gives x = 1 and 2x = 2 which gives x = 1. Hence f(2x) will have x intercepts at x = 1 and x = 1. The yintercept is at y = 2 since f(0) = 2 and f(2*0) = f(0) = 2.
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 since f^{ 1}(1) = 0 , f(0) = 1 , and since f^{ 1}(0) = 2 , f(2) = 0
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