983.The graph of the function f(x) = x^82 – – 6 x 4 has an
open hole at x= 8.
a.The graph actually has a vertical asymptote
at x= 8 because this value makes the denom-
inator equal to zero.
b.The graph actually has a horizontal asymp-
tote at x= 8 because this value makes the
denominator equal to zero.
c.There is no error.
984.Iff(2) = 5, then the point (5, 2) must be on the
graph ofy= f(x).
a.The coordinates of the point that is known
to lie on the graph ofy= f(x) are reversed; it
should be (2,5).
b.The given information is insufficient to
make any conclusion about a point being on
the graph ofy= f(x). All that can be said is
that 2 is in the range off.
c.There is no error.
985.The range of the function f(x) = (x– 1)^2
is [0,∞).
a.The graph offis the graph ofg(x) = x^2
shifted vertically up one unit. Since the range
ofgis [0,∞), it follows that the range off
must be [1,∞).
b.The graph offis the graph ofg(x) = x^2
shifted vertically down one unit. Since the
range ofgis [0,∞), it follows that the range
offmust be [–1,∞).
c.There is no error.
- Iff(x) = 5 and g(x) = x, it follows that
(f ̊ g)(–2) = 5.
a.The composition was computed in the
wrong order. The correct output should
be 5 .
b.–2 is not in the domain ofg,so that the
composition is not defined at –2.
c.There is no error.
987.The x–intercept off(x) = x^3 + 1 is (0, 1).
a.The point (0,1) is the y–intercept off,not
the x–intercept.
b.There are no x–intercepts for this function
because x^3 + 1 is always positive.
c.There is no error.988.The graph ofg(x) = 2–xis increasing as x moves
from left to right through the domain.
a.The graph ofg is actually decreasing as x
moves from left to right through the
domain.
b.There are intervals on which the graph ofg is
increasing and others on which it is
decreasing.
c.There is no error.989.The graph ofy= f(x+ 3) is obtained by shifting
the graph ofy = f(x) to the right 3 units.
a.The graph ofy= f(x+ 3) is actually obtained
by shifting the graph ofy= f(x) to the left 3
units.
b.The graph ofy= f(x+ 3) is actually obtained
by shifting the graph ofy= f(x)vertically up
3 units.
c.There is no error.- COMMON ALGEBRA ERRORS–