\[ \mathbf{18}.s=\sqrt{A}\quad\mathbf{19}.P=4\sqrt{A}\quad\mathbf{20}.A=P^{2}/16\quad\mathbf{21}.A=C^{2}/4\pi\quad\mathbf{22}.C=2\sqrt{\pi A}\quad\mathbf{23}.A=\frac{\pi P^{2}}{2(\pi+2)^{2}} \]
- $ P = (+ 2) $ 25. $ d = c $ (where c is the cube’s face diagonal, and d its body diagonal)
\[ \mathbf{26.}\;V=(S/6)^{3/2}\;\mathbf{27.}\;V=\frac{\pi}{6}d^{3}\quad\mathbf{28.}\;r=\sqrt[3]{\frac{3V}{4\pi}}\quad\mathbf{29.}\;V=A^{3/2}/6\sqrt{\pi}\quad\mathbf{30.}\;S=\sqrt{3}x^{2}\;\mathbf{31.}\;x=\sqrt{\frac{S}{5\sqrt{3}}}\quad\mathbf{32.}\;A=\frac{h^{2}}{\sqrt{3}} \]
- $ (2b + a, f(2b + a)) $ b) $ $ c) $ y = ()x $
\[ \mathbf{34.a)\left(\frac{1}{2}-\sqrt{2},0\right)and\left(\frac{1}{2}+\sqrt{2},0\right)\qquad b)\left(0,-\frac{7}{8}\right)\qquad c)\left(2,\frac{1}{8}\right)\qquad d)\left(\frac{1}{2}-\sqrt{6},2\right)and\left(\frac{1}{2}+\sqrt{6},2\right)} \]
\[ e)\left(\frac{2+\sqrt{19}}{2},\frac{6+\sqrt{19}}{4}\right)and\left(\frac{2-\sqrt{19}}{2},\frac{6-\sqrt{19}}{4}\right)\quad f)\left(\frac{1-\sqrt{22}}{6},\frac{2\sqrt{22}-23}{36}\right)and\left(\frac{1+\sqrt{22}}{6},\frac{-23-2\sqrt{22}}{36}\right) \]
\[ g)slope=-1/3\qquad h)No\qquad i)\left(3/2,-1/2\right) \]
- You’ve learned how to find the solutions of quadratic equations, but not of cubic equations, such as the one that emerges in this problem. There is indeed a “cubic formula” analogous to the quadratic formula, but it is far too gory for a family-friendly book like this one.
Intersections at (0,1) and (612/983,0) 37. a) yes for both b) yes for the left curve, no for the right
- 3 b) $ 128’ $ , $ 80’ $ c) $ /2 $ sec d) $ $ sec e) $ 16’11’’ $
- $ P: (x, f(x)) $ , $ Q: (x + h, f(x + h)) $ b) $ $
- Upper, lower halves of the unit circle. c) $ c = 2 $
- $ m(x)= $ . Domain: $ -1 x $ . Range: $ |m(x)| $ . (Think geometrically, and you’ll see this.)
- Writing y as a function of x means rewriting the equation in an equivalent form with y isolated on one side. Equivalent equations have the same graph (since they have the same solutions), so if we could write y as a function of x, the function’s graph would be that of the original equation. But… this cannot be: the original’s graph is cut multiple times by the same vertical line (as a graphing program reveals), which cannot happen in the graph of a function. Hence, there’s no way to rewrite the original equation so that y is a function of x.
It is remarkable that we were able to answer a purely algebraic question (can such-and-such an equation be manipulated into such-and-such a form?) with an essentially geometric argument. Such is the power of coordinate geometry.
Chapter 6
Check your graphs with a computer graphing program.
4, $ $ 7. Yes, in this case, a h-stretch by a factor of $ $ corresponds to a v-stretch by a factor of 2.
No. Any v-stretch will move the point $ (0,1) $ somewhere else, so the resulting graph could not coincide with the h-stretched graph, which includes that point.
Topmost: (1,1); Leftmost: (1/2,0) 14. a) 6 distinct orders… b) …but only 4 distinct graphs.
- sub. \((x-8)\) for \(x\) b) sub. \((x+8)\) for \(x\) c) sub. \((y+8)\) for \(y\) d) sub. \((y-8)\) for \(y\)
sub. $ (x + 1/2) $ for x f) sub $ (y/6) $ for y g) sub. $ (3y) $ for y h) sub. $ (x - 4) $ for x
sub. $ (x/10) $ for x j) sub. $ (10x) $ for x k) sub. $ (x/17) $ for x l) sub. $ (y - 2) $ for y
sub. $ (2/3)y $ for y n) sub. $ (16/3)y $ for y o) sub. -y for y
sub. $ (17/2)y $ for y r) sub. -x for x s) sub. $ (y + 1) $ for y
\[ \mathbf{16}.y=\frac{1}{x-3}\quad\mathbf{17}.x^{2}+\frac{(y+3)^{2}}{4}=1\quad\mathbf{18}.(x+1)^{2}+(y-1)^{2}=2 \]