- a.Look for the selection that has a negative x-
coordinate and positive y-coordinate. Since
a0, it follows that –a0. The choice that
lies in Quadrant II is (–a,a).
- d.Note that ifxis a negative integer, then –x^3 =
–(x)(x)(x) must be positive (because it is a
product of an even number of negative inte-
gers). Likewise, since xand yare both negative
integers,xy^2 is negative (because it is a product
of an odd number of negative integers). Hence,
the x-coordinate of (–x^3 ,xy^2 ) is positive and its
y-coordinate is negative. So, the point lies in
Quadrant IV.
- b.Using the fact that xand yare both assumed
to be negative integers, we must determine the
signs of the coordinates of the point ( ,x^1 y).
To this end, note that –x^2 is negative, (–y)^3 is
positive (since –yis a positive integer and the
cubes of positive integers are positive integers),
and xyis positive (since it is a product of an
even number of negative integers). The x-
coordinate of the given point is therefore neg-
ative (since the numerator is negative and
denominator is positive, thereby creating a
quotient involving an odd number of negative
integers) and the y-coordinate is positive. So,
the point lies in Quadrant II.
- c.Since the y-coordinate of the point (–x, –2)
is –2, it follows that for any real number x, the
point must lie somewhere strictly below the x-
axis. Ifx0, the x-coordinate of the point is
negative, so that it lies in Quadrant III, while it
lies in Quadrant IV ifx0 and on the y-axis
ifx= 0. So, the correct choice is c.
- d.The phrase “yis nonpositive” can be expressed
symbolically as y 0. As such, –y 0. Since
the x-coordinate of the point (1, –y) is positive
and the y-coordinate is not negative, the point
must be in Quadrant I or on the
x-axis. It can be onthe x-axis ify= 0. Neither
anor bis true.
Set 16(Page 28)
- a.Convert the given equation 3y– x= 9 into
slope-intercept form by solving for y, as follows:
3 y– x= 9
3 y= x+ 9
y= ^13 x+ 3
The slope of this line is the coefficient ofx,
namely ^13 .
- b.The line whose equation is y= –3 is hori-
zontal. Any two distinct points on the line
share the same y-value, but have different
x-values. So, computing the slope as “change
in yover change in x” results in 0, no matter
which two points are used.
- a.Convert the given equation 8y = 16x– 4
into slope-intercept form by solving for y:
8 y = 16x– 4
y = 2x– ^12
So, the y-intercept is (0,–^12 ).
- d.Substituting x= 3 and y= 1 into the equa-
tion y= ^23 x– 1 yields the true statement 1 =
^23 (3) – 1, which implies that the point (3,1) is
on this line.
- b.The slope-intercept form of a line with
slope m and y-intercept (0,b) is y= mx+ b.
So, the equation of the line with slope –3 and
y-intercept of (0,2) is y= 3x+ 2.
- a.First, choose two of the five points listed and
compute the slope. We will use the first two
listed, (1,7) and (2,10). The slope is m= ^120 – – 17
= 3. Next, use one of the points, such as (1,7),
and the slope m= 3 to write the equation of
–x^2
(–y)^3
ANSWERS & EXPLANATIONS–
