10) The y- intercept of a graph occurs when x = 0, so we get x = −1. Analogously,
for the x − intercept, we set y = 0 and get 0 = y2 − 1, so y2
= 1 and y = ±1. For
symmetry, −x = −(y2) + 1 ≠ y2 − 1, so there is no
symmetry about the y-axis.
This is also not equal to (−y)2 − 1, so we don’t have symmetry about
either. However, we do have that (−y)2 −1 = y2 −1 = x, so
we get that the graph
is symmetric about the x-axis. The graph looks like a sideways parabola with the
12) so we can see that there are no
intercepts at all, as x being 0 would
lead to us dividing by it and while y will approach 0, it cannot actually
it. For symmetry, not that so it is symmetric
about the y-axis, and
it has the coordinate axes as asymptotes.
18) x = 0 implies y3 = 1, so y = 1, and y = 0 implies that the
−1. It’s easily verified that there are no symmetries around the origin, x-axis,
24a) x = 0 gives |y| = 2, so y = ±2, and y = 0 gives x = 2. There is symmetry
about the x-axis.
40) The center is (−4,−2), and radius is
Plugging in (0, 1), we get 42 + 32 =
16 + 9 = 25 ≠0, so the point is not on the line .
44) Completing the square, we get (x − 5)2 + (y + 1)2 =
−17 + 25 + 1 = 9, so
the center is (5,−1) and radius is 3. To see where any y-intercept is, plug in x
and we get 25+(y+1)2 = 9 or (y+1)2 = −16, which is
impossible, so we conclude
there are no y-intercepts.
48) The first step here is to get the coefficients of x2 and y2
to be 1, so we divide
through by 3, obtaining Completing the
square, we get
So the center is atand
the radius is
To find any y-intercepts, plug in x = 0, and
hence and finally,
If you had any issues with the process of completing the square in the problems
above, please see me and we can work some more examples.
60a) The y-intercept in both is easily found to be 8, since in the first
we get y = 8 and in the second y = |8| = 8 when we set x = 0. For x-intercepts,
set y = 0 in the first and get 0 = x2 −6x+8 = (x−4)(x−2), so they are
x = 4, 2,
and in the second, we get 0 = |(x−4)(x−2)| which produces the same x-intercepts.
60b) The identical portions are (−∞, 2] and [2,∞).
60c) The curves are different over the interval (4, 2). The second curve,
is just a reflection across the x-axis of this region away from being the first.
should make sense, since the absolute value cannot ever be negative, so where
original equation goes negative, the absolute value gives its reflection.
36) Let’s begin by assuming a = 1; if need be, we can multiply through later
to clear out any fractions . Then b = −(r1 + r2) and c = r1r2.
So b = −4 and
c = −1, hence the quadratic equation with these roots is x2 − 4x − 1
38) Again, begin by assuming a = 1. Then b =
and c = so clearing
out the denominators so that all coefficients are integers, we get 9x2−24x+11
46) b = −28, a = 4, and c = 49, so b2 = 784 and 4ac = 784, hence the
is 0 and there is only onereal root with multiplicity 2.
54) Here we have a = b = k and c = 1, so the discriminant is k2 − 4k.
this equal to 0 and solving, we get that k = 0, 4.
66) As above, we want to find the ex pression for the discriminant , set it equal
to 0, and solve. So b = 2(k + 1), a = 1, and c = k2, hence b2
= 4k2 + 8k + 4 and
4ac = 4k2, and so the discriminant is 8k + 4. Setting equal to 0 and