Math Homework 2 Solutions

1.7
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 the origin,
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
appropriate intercepts.

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 intersect
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 x-intercept is
−1. It’s easily verified that there are no symmetries around the origin, x-axis, or
y-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 = 0,
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 x ² 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 we get
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 equation
we get y = 8 and in the second y = |8| = 8 when we set x = 0. For x-intercepts, we
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, however,
is just a reflection across the x-axis of this region away from being the first. This
should make sense, since the absolute value cannot ever be negative, so where the
original equation goes negative, the absolute value gives its reflection.

2.1

6) The quadratic formula gives us that the roots


 Again, we use the quadratic formula :


and so this root has multiplicity 2.

32) The equation is so the sum of the roots is and
the product is 5

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 = 0.

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 = 0.

46) b = −28, a = 4, and c = 49, so b² = 784 and 4ac = 784, hence the discriminant
is 0 and there is only one real root with multiplicity 2.

54) Here we have a = b = k and c = 1, so the discriminant is k² − 4k. Setting
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 solving gives
k =