1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1

  1. b.First, note that x^2 + 1 does not factor, so x= 1
    is a vertical asymptote for the graph off. Since
    the degree of the numerator of the fraction is
    exactly one more than that of the denominator,
    we can conclude that the graph has no horizon-
    tal asymptote, but does have an oblique asymp-
    tote. Hence, II is a characteristic of the graph
    off.


Next, while there is a y-intercept, (0,3), there is
no x-intercept. To see this, we must consider
the equation,f(x) = 0 which is equivalent to

2 – xx

2





+
1
^1 = = –(x^2
x









2 x
1
+3)= 0.

The x-values that satisfy such an equation are
those that make the numerator equal to zero
and do not make the denominator equal to
zero. Since the numerator does not factor, we
know that (x– 1) is not a factor of it, so we
need only solve the equation.x^2 – 2x+ 3 = 0.
Using the quadratic formula yields

x= = = 1 ±i 2 

Since the solutions are imaginary, we conclude
that there are no x-intercepts. Hence, III is not
a characteristic of the graph off.


  1. b.The expression for fcan be simplified as
    follows:


= =

= x+x^3

Since x= 2 makes both the numerator and
denominator of the unsimplified expression
equal to zero, there is a hole in the graph offat
this value. So, statement I holds.

Next, since the degrees of the numerator and
denominator are equal, there is a horizontal
asymptote given by y= 1 (since the quotient of
the coefficients of the terms of highest degree
in the numerator and denominator is 11 =
1). Because x= 0 makes the denominator equal
to zero, but does not make the numerator equal
to zero, it is a vertical asymptote. So, statement
II holds.

Finally, since x= 0 is a vertical asymptote, the
graph offcannot intersect it, and there is no y-
intercept. So, statement III does not hold.


  1. c.The y-values off(x) = ^1 xget smaller as x-
    values move from left to right. The graph is as
    follows:


–10 –8 –6 –4 –2 2 4 6 8 10
–2
–4
–6
–8
–10

10
8
6
4
2

(x– 2)^2 (x+ 3)
x(x– 2)^2

(–(x– 2))^2 (x+ 3)
x(x– 2)^2
(2 –x)^2 (x+3)
x(x– 2)^2

2±–8
2
–(–2) ±(–2)^2 – 4(1)(3)
2(1)

2(x– 1) – (x^2 + 1)
x– 1

ANSWERS & EXPLANATIONS–
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