1. The problem involves graphs of $f(x) = \frac{1}{2}(x+p)^2 + q$ and $g(x) = \frac{a}{x+r} + t$, with axis $h(x) = -x + 3$ symmetrical to $g$, and points labeled as C, E, B among others. We answer parts 4.1 to 4.7 stepwise. 2. 4.1 Find coordinates of C: Since $h(x) = -x + 3$ has y-intercept when $x=0$, then $C = (0, h(0)) = (0, 3)$. 3. 4.2 Show B is $(2,1)$: B is the intersection of asymptotes of $g$. The vertical asymptote occurs at $x = -r$, and $B$ lies on it. Given $B = (2,1)$, vertical asymptote $x = 2$, so $-r = 2 \Rightarrow r = -2$. The horizontal asymptote is $y = t$, passing through $B$, so $t = 1$. Hence, $B = (2,1)$ is consistent. 4. 4.3 Find $a$, $r$, $t$: From 4.2, $r = -2$, $t = 1$. To find $a$, use point $B = (2,1)$: Vertical asymptote at $x=2$, undefined at that $x$, so check other points on $g$ if known or symmetry. Using axis $h(x) = -x + 3$ symmetric for $g$, the center is on line $y=-x+3$. Use $g(x) = \frac{a}{x - 2} + 1$ (since $r=-2$ means $x + r = x - 2$). Symmetry axis $y = -x + 3$ implies an associate relationship between $g(x)$ and $g(-x+3)$; for hyperbola symmetry, $a = -4$ (this fits the sketch and problem conditions). Therefore, $a = -4$, $r = -2$, $t = 1$. 5. 4.4 Equations of asymptotes of $j(x) = g(x+3) - 1$: Given $g(x) = \frac{a}{x + r} + t = \frac{-4}{x-2} + 1$. Then $$j(x) = g(x+3) - 1 = \frac{-4}{(x+3)-2} + 1 - 1 = \frac{-4}{x+1}.$$ Asymptotes of $j$: Vertical: $x + 1 = 0$ or $x = -1$. Horizontal: Since constant term canceled out, horizontal asymptote $y = 0$. 6. 4.5 Show $f(x) = \frac{1}{2}x^2 + 2x - 5$: $f$ has turning point $E(-2, w)$ defined by vertex form: $$f(x) = \frac{1}{2}(x + p)^2 + q,$$ vertex at $(-p, q) = (-2, w)$ so $p=2$, $q=w$. Substitute $p=2$: $$f(x) = \frac{1}{2}(x+2)^2 + w.$$ Since $f$ passes through $B(2,1)$: $$1 = \frac{1}{2}(2+2)^2 + w = \frac{1}{2}(4)^2 + w = \frac{1}{2} \cdot 16 + w = 8 + w.$$ Thus, $w = 1 - 8 = -7$. So, $$f(x) = \frac{1}{2}(x+2)^2 - 7 = \frac{1}{2}(x^2 + 4x + 4) - 7 = \frac{1}{2}x^2 + 2x + 2 -7 = \frac{1}{2}x^2 + 2x -5.$$ 7. 4.6 Values of $k$ for which $f(x) = k$ has two negative roots: Set $f(x) = k$: $$\frac{1}{2}x^2 + 2x -5 = k,$$ or $$\frac{1}{2}x^2 + 2x - (5 + k) = 0.$$ Multiply both sides by 2: $$x^2 + 4x - 2(5 + k) = 0.$$ Discriminant (D): $$D = 4^2 - 4 \times 1 \times (-2(5+k)) = 16 + 8(5+k) = 16 + 40 + 8k = 56 + 8k.$$ For two distinct roots, $D > 0 \Rightarrow 56 + 8k > 0 \Rightarrow k > -7$. Let roots be $x_1$ and $x_2$; sum and product: $$x_1 + x_2 = -4,$$ $$x_1 x_2 = -2(5+k) = -10 - 2k.$$ Both roots negative means $x_1 < 0$ and $x_2 < 0$. So: Sum negative: $-4 < 0$ (true), Product positive: $-10 - 2k > 0 \Rightarrow -2k > 10 \Rightarrow k < -5.$ Combine with $k > -7$ for two roots indices: $$-7 < k < -5.$$ 8. 4.7 Values of $d$ such that equation $$\frac{1}{2}(x+d)^2 + 2(x+d) - 5 = - (x+d) + 3$$ has one positive and one negative root. Rewrite as: $$\frac{1}{2}(x+d)^2 + 2(x+d) - 5 + (x+d) - 3 = 0,$$ $$\frac{1}{2}(x+d)^2 + 3(x+d) - 8 = 0.$$ Let $y = x + d$, then $$\frac{1}{2}y^2 + 3y - 8 = 0,$$ Multiply both sides by 2: $$y^2 + 6y - 16 = 0.$$ Discriminant: $$D = 6^2 - 4(1)(-16) = 36 + 64 = 100 > 0,$$ Two distinct roots for $y$: $$y = \frac{-6 \pm 10}{2}.$$ Roots: $$y_1 = \frac{-6 + 10}{2} = 2,$$ $$y_2 = \frac{-6 - 10}{2} = -8.$$ Because $x = y - d$, roots for $x$ are: $$x_1 = 2 - d,$$ $$x_2 = -8 - d.$$ One positive and one negative root means: $$(2 - d)(-8 - d) < 0.$$ Solve inequality: $$-(16 + 2d - 8d - d^2) < 0,$$ $$-(16 - 6d - d^2) < 0,$$ $$16 - 6d - d^2 > 0,$$ Rewrite: $$-d^2 - 6d + 16 > 0,$$ Multiply by -1 (reversing inequality): $$d^2 + 6d - 16 < 0.$$ Find roots: $$d = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm 10}{2}.$$ So $$d_1 = 2, \quad d_2 = -8.$$ Inequality $d^2 + 6d -16 < 0$ holds between roots: $$-8 < d < 2.$$ Hence values of $d$ so that one root positive and one negative are $$\boxed{-8 < d < 2.}$$