College Algebra Problems With Answers
sample 4 : Graphs of Functions

A set of college algebra problems on graphs of functions with answers, are presented. The questions are related to the transformations of graphs, graphs of inverse functions, symmetry of graphs, reading values from graphs, find domain and range and intervals on increase and decrease. The solutions are at the bottom of the page.

Problems

Problem 1

The graph of \( f(x) \) is shown below. Draw the graph of \( y = - f(x - 3) - 2 \)
college algebra problem 1, function f(x)

Figure 1. Graph of Function \( f \) - Problem 1

Problem 2

Complete the graph given below so that it is symmetric with respect to the origin.
college algebra problem 2, function f(x).

Figure 2. Graph of Problem 2

Problem 3

The graph of \( f(x) \) is shown below. Sketch the graph of \( y = f(-x + 1) - 1 \) (hint: see Graphing by Translation, Scaling and Reflection)
college algebra problem 3, function f(x).

Figure 3. Graph of Function \( f \) - Problem 3

Problem 4

The graph of \( f(x) \) is shown below. Sketch the graph of the inverse of \( f \).
college algebra problem 4, function f(x).

Figure 4. Graph of Function \( f \) - Problem 4

Problem 5

The graph of \( h(x) \) is shown below.
college algebra problem 5, function f(x).

Figure 5. Graph of Function \( f \) - Problem 5

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.

Problem 6

The graph of \( f(x) \) is shown below. Sketch the graph of f(2x).
college algebra problem 6, function f(x).

Figure 6. Graph of Function \( f \) - Problem 6

Problem 7

The graph of \( f^{-1}(x) \) is shown below.
Evaluate the following: \( f(0) \) , \( f(2) \)
college algebra problem 7, function f(x).

Figure 7. Graph of Function \( f \) - Problem 7

Solutions to the Above Problems

Solution to Problem 1

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 x-axis : \( - f(x - 3) \)
3 - shift down 2 units : \( - f(x - 3) - 2 \)
4 - connect the transformed points to sketch the transformed graph shown in red below.
college algebra, Graph of Solution to Problem 1.

Figure 8. Graph of Solution to Problem 1

Solution to Problem 2

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 symmetric with respect to the origin.(see the whole graph black and red below).
college algebra, Graph of Solution to Problem 2.

Figure 9. Graph of Solution to Problem 2

Solution to Problem 3

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.
college algebra, Graph of Solution to Problem 3.

Figure 10. Graph of Solution to Problem 3

Solution to Problem 4

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).
college algebra, Graph of Solution to Problem 4.

Figure 11. Graph of Solution to Problem 4

Solution to Problem 5

\( h(-2) + h(2) = -3 + 1 = -2 \)
Domain: \([-5 , 5]\)
Range: {-3} U \([0 , 5]\)
increase: \([-3 , 2)\) , \([2 , 5]\)
decrease: \([-5 , -3]\)
constant: \([-2 , 2)\)

Solution to Problem 6

Function \( f \) has two x-intercepts: \( x = 2 \) and \( x = -2 \) . \( f(2x) \) will also have x-intercepts 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 y-intercept is at \( y = 2 \) since \( f(0) = 2 \) and \( f(2*0) = f(0) = 2 \).
college algebra, Graph of Solution to Problem 6.

Figure 12. Graph of Solution to Problem 6

Solution to Problem 7

since \( f^{-1}(1) = 0 \) , \( f(0) = 1 \) , and since \( f^{-1}(0) = 2 \) , \( f(2) = 0 \)

More References and links

Graphing Mathematical Functions
Inverse Function
Graphing by Translation, Scaling and Reflection
Algebra Questions and problems
More ACT, SAT and Compass practice