Chapter 4

    1. No to $ (-4,2) $ , Yes to $ (1,1) $
  1. No: If x were 0 for some point on the graph, then we’d have 0 = 1, which is nonsense.

\[ \begin{pmatrix}0,-4\end{pmatrix}\quad\mathbf{3}.\quad-10\quad\mathbf{4}.\begin{pmatrix}0,\frac{9}{2}\end{pmatrix},\begin{pmatrix}0,-3\end{pmatrix}\quad\mathbf{5}.\begin{pmatrix}1,1\pm\sqrt{11}\end{pmatrix} \]

\[ x=-4\quad b)y=2\quad c)y=0\quad d)x=0\quad e)x=a\quad f)y=a\quad g)x=0 \]

    1. 10.25 b) 11.75 c) $ T = -20 $ d) $ x = 5 $ , $ y = 9 $ , $ = $ e) $ x = -5 $ , $ y = -9 $ , $ = $

  2. $ - $ 15. The line’s slope will be negative.

    1. y = 2x - 9, which crosses the x-axis at $ (9/2, 0) $ and the y-axis at $ (0, -9) $ .

  1. $ y = -6x + 17 $ , which crosses the x-axis at $ (17/6, 0) $ and the y-axis at $ (0, 17) $ .

  2. $ y = -(3/2)x + (5/6) $ , which crosses the x-axis at $ (5/9, 0) $ and the y-axis at $ (0, 5/6) $ .

  3. $ y = ( ) x - ( ) $ , which crosses the x-axis at $ ( , 0 ) $ and the y-axis at $ ( 0, - ) $ .

  4. $ y = (8/35)x + (12/35) $ , which crosses the x-axis at $ (-3/2, 0) $ and the y-axis at $ (0, 12/35) $ .

  5. $ y = mx + b $ , which crosses the x-axis at $ (-b/m, 0) $ and the y-axis at $ (0, b) $ .

\[ \textcircled{d}y=7x+2\qquad\textcircled{b}y=-(2/3)x+8/3\qquad\textcircled{c}y=x-7/3\qquad\textcircled{d}y=7x-14 \]

    1. $ 1.8^{} $ F b) $ 5(1.8^{}) = 9^{} $ F c) $ C = (5/9)(F - 32) $ d) Yes. $ -40^{} $ F = $ -40^{} $ C.

  2. [Use a graphing program to check your solutions.]

    1. $ y = (2/5)x + 2 $ b) $ y = -(1/4)x + (3/4) $ c) y = -2x

\[ \mathbf{21.a)y=-4x+14\quad b)y=-(2/5)x\quad\mathbf{22.a)y=(1/4)x+(5/4)\quad b)y=(5/2)x} \]

  1. $ 4 $ inches 25. $ $ miles 26. No: the PT does not apply here, unless this happens to be a right triangle.

  2. Yes: since the hypotenuse of a right triangle is always its longest side, the hypotenuse here must be

6 units (since $ 6 > 2 $ ). Accordingly, the shortest side will be, by the PT, 4 units long.

\[ \mathbf{30.a)10}\quad\mathbf{b)\sqrt{109}}\quad\mathbf{c)\sqrt{1129}/4}\quad\mathbf{d)\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}}\quad\mathbf{e)8} \]

  1. This happens when the segment itself is parallel to an axis. Thus, strictly speaking, the proof I gave in the text is not quite complete: It establishes the distance formula for all pairs of points except those that lie on the same vertical or horizontal line. Fortunately, the distance formula still holds in such cases, but its validity must be established there separately. This is easy. If two points lie on the same horizontal line, the distance between them is obviously $ |x| $ . (If this isn’t obvious to you, play around with specific examples until it is.) It remains only to verify that the distance formula actually produces this result under these circumstances. It does: For any two points on the same horizontal line, $ y = 0 $ , so the distance formula produces the result.

\[ d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{(\Delta x)^{2}}=|\Delta x|, \]

which is correct. Thus, the distance formula holds even when the two points lie on the same horizontal line. (A similar argument takes care of the case when the points lie on the same vertical line, as you should verify.) The moral of the story is that the distance formula holds under all circumstances, though crossing every t and dotting every i in the proof would make it very tedious to read, obscuring the proof’s essential simplicity.

\[ \mathbf{31.}\ x^{2}+y^{2}=1\qquad\mathbf{32.}\ a)\ (x+3)^{2}+(y-2)^{2}=16\qquad\mathbf{b})\ (0,2\pm\sqrt{7}),\ (-3\pm2\sqrt{3},\ 0)\qquad\mathbf{c})\ (-1,\ 2\pm2\sqrt{3}) \]

    1. $ (x-1){2}+(y-1/2){2}=1/4 $ b) $ (1,0) $ c) $ (1/2,1/2) $ , $ (3/2,1/2) $
    1. $ (x - 1/2)^{2} + (y - 5)^{2} = 29/4 $ . Intersections w/ axes: $ (0, 5 ) $

  1. $ (x-3){2}+(y+1){2}=37 $ . Intersections w/ axes: $ (0,-1) $ , $ (9,0) $ , $ (-3,0) $ .

  2. $ x^{2} + y^{2} = 49/4 $ . Intersections w/ axes: $ (0, /2) $ , $ (/2, 0) $ .

  1. $ (x-1/2){2}+(y-1/2){2}=1/2 $

    1. $ 40 $ feet b) $ 20 $ feet c) $ 1600 $ feet $ ^{2} $ .
    1. Since the full circle’s center is $ (2, -3) $ , the points on its upper half all have y-coordinates of $ (-3) $ or greater. Thus, the circle’s upper half corresponds to $ y = -3 + $ , where y is $ (-3) $ plus something else. (The “something else”, the radical term, is always positive or zero.) The circle’s lower half has a similar story.

  1. The left and right halves of the circle c) $ y = - $ d) $ x = -1 + $

  2. Bottom half of the circle centered at the origin with radius 2.

  1. (100, 10) 41. Negatives lack real square roots, but they have cube roots. 45. (1, 1)

  2. The graph of $ y = |x| $ takes half of the graph of y = x and marries it to half the graph of y = -x.

This makes sense because $ |x| $ is equal to x when x is positive, and equal to -x when x is negative.

    1. (1, 1) b) $ (, ) $ c) $ (, ) $ , $ (, ) $ d) $ (, ) $ , $ (, ) $

\[ \begin{aligned}e)\left(0,0\right),\left(\sqrt{3},3\sqrt{3}\right),\left(-\sqrt{3},-3\sqrt{3}\right)\quad&f)\left(1+\sqrt{2},3+2\sqrt{2}\right),\left(1-\sqrt{2},3-2\sqrt{2}\right)\end{aligned} \]

\[ \mathsf{g})\left(\sqrt{\frac{\sqrt{5}-1}{2}},\frac{\sqrt{5}-1}{2}\right),\left(-\sqrt{\frac{\sqrt{5}-1}{2}},\frac{\sqrt{5}-1}{2}\right)\quad\mathsf{h})\left(0,0\right),\left(2\sqrt{2},\sqrt{2}\right),\left(-2\sqrt{2},-\sqrt{2}\right)\quad\mathsf{i})\left(4000000,2000\right) \]

\[ \mathrm{j})\left(\sqrt{\frac{7}{2}},1+\sqrt{\frac{7}{2}}\right) \]

  1. The lines do not intersect. 50. The line and circle do not intersect.

    1. No: Their slopes differ. b) $ y = -x $ c) $ (, -) $

  3. \(\left(\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}\right),\left(\sqrt{2-\sqrt{3}},\sqrt{2+\sqrt{3}}\right),\left(-\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}\right),\left(-\sqrt{2-\sqrt{3}},-\sqrt{2+\sqrt{3}}\right)\)

[Note: These coordinates could appear in other equivalent forms, too.]

\[ \mathbf{53.a)}\left(x-\frac{3}{2}\right)^{2}+(y-1)^{2}=\frac{13}{4}\qquad b)(x-2)^{2}+(y-4)^{2}=20\qquad c)(x-3)^{2}+(y-1)^{2}=10 \]

  1. Using 7 and 3 would work well; this would let us eliminate x when we add the resulting equations.

\[ \mathbf{55.}\left(6,-9\right)\qquad\mathbf{56.}\;a)\left(13/42\;,\;3/14\right)\quad\mathbf{b)}\left(9/5,-1/5\right)\quad\mathbf{c)}\left(1/6,-5/18\right)\quad\mathbf{d)}\left(2/3,\;1/3\right) \]

Chapter 5

  1. 144 2. $ h(3) = 144 $

    1. The area of an equilateral triangle whose sides are 10 units long.
  1. It has one solution, which represents the side length of an equilateral triangle whose area is $ 10 , units^{2} $ .
    1. 29/4 b) 15 c) 2 d) 145/64 e) All real numbers f) Yes, since $ k(4) = 5 $ . g) 1 h) There isn’t one.
    1. $ k(0) = 1 $ b) $ k(10) = 26 $ c) $ k(1) = 5/4 $ 6. Yes

  3. No, since isocles triangles with the same base can have different areas 8. Yes 9. a) 26 b) 7

    1. All real numbers b) No c) x = 0, 1 d) 7 e) 13 f) 79/49 g) $ 9 + $ h) 31
    1. All real numbers but 0 b) Yes, because $ k ( 5 + ) = 10 $ c) $ x = $ d) $ g(1) = 0 $
  1. $ - $ f) $ $ g) $ $ h) $ $
    1. x = 1, 3/4 b) 21 c) -3 d) $ 2x^{2} - 13x + 21 $ e) $ 2x^{2} - x - 3 $
  1. $ 2x^{3} - 7x^{2} + 3x $ g) $ 2x^{2} - 3 $ h) 1 i) 0
    1. All real numbers 3 or greater b) All reals except -2 c) All reals except 0 and 2 d) All reals -1/4 or greater, except 0 and 1.

\[ \mathbf{17}.\ a)\ 1\quad\mathbf{b)\ 2x+h}\quad\mathbf{c)\ -2x-h}\quad\mathbf{d)\ 3x^{2}+3hx+h^{2}}\quad\mathbf{e)\ -2x+3-h}\quad\mathbf{f)\ \frac{2}{x^{2}+xh}}\quad\mathbf{g)\ \frac{1}{\sqrt{x+h}+\sqrt{x}} \]