- COMMON ALGEBRA ERRORS–
990.The graph ofy= f(x) – 2 is obtained by shifting
the graph ofy= f(x)down 2 units.
a.The graph ofy= f(x) – 2 is obtained by shift-
ing the graph ofy= f(x) to the left 2 units.
b.The graph ofy= f(x) – 2 is obtained by shift-
ing the graph ofy= f(x) to the right 2 units.
c.There is no error.
991.If,f(x) = x^4 , then f(x– h) = f(x) – f(h) = x^4 – h^4.
a.You cannot distribute a function across parts
of a single input. As such, the correct state-
ment should be f(x– h) = (x– h)^4.
b.The second equality is incorrect because
you must also square the –1. As such, the
correct statement should be f(x– h) =
f(x) – f(h) = x^4 + h^4.
c.There is no error.992.The graph ofy= 5 does not represent a function
because it does not pass the horizontal line test.
a.The graph ofy= 5 passes the vertical line
test, so it represents a function. It is, how-
ever, not invertible.
b.The fact that y= 5 does not pass the hori-
zontal line test does not imply it is not a func-
tion. However, since the range of a function
must consist of more than a single value, we
conclude that it y= 5 cannot represent a
function.
c.There is no error.Set 64 (Answers begin on page 276)
This problem set highlights common errors made when dealing with linear systems of equations and matrix algebra.
993.The system has infinitely many solutions.
a.Since adding the two equations results in the false statement 0 = 8, there can be no solution of this
system.
b.The slopes of the two lines comprising the system are negatives of each other. As such, the lines are
perpendicular, so the system has a unique solution.
c.There is no error.994.The system has no solutions.
a.Since multiplying the first equation by –2 and then adding the two equations results in the true state-
ment 0 = 0, there are infinitely many solutions of this system.
b.The two lines comprising the system intersect, so the system has a unique solution.
c.There is no error.410 2
xy
xy25 1––
––
=
* =
xy
xy