1.1.1 Show that equation of $f(x)$: Given points $B(-5,0)$ and $C(1,0)$ are roots of $f(x)=ax^2+bx+c$. We write $f(x)=a(x+5)(x-1)$. Point $A(-1,4)$ lies on the curve, so: $$4=a(-1+5)(-1-1)=a(4)(-2)=-8a\implies a=-\frac{1}{2}.$$ Then: $$f(x)=-\frac{1}{2}(x+5)(x-1)=-\frac{1}{2}(x^2+4x-5)=-\frac{1}{2}x^2 -2x + \frac{5}{2}.$$ 1.1.2 Determine turning point of $f(x)$: Form $f(x)=-\frac{1}{2}x^2 - 2x + \frac{5}{2}$; vertex $x=-\frac{b}{2a}=-\frac{-2}{2(-\frac{1}{2})}=-\frac{-2}{-1}=2.$ Calculate $f(2)$: $$f(2) = -\frac{1}{2}(2)^2 - 2(2) + \frac{5}{2} = -2 -4 + 2.5 = -3.5.$$ Turning point is at $(2, -3.5)$. 1.1.3 Determine equation of $g(x)$ (line through A and C with $y=mx+c$): Slope $m=\frac{0-4}{1-(-1)}=\frac{-4}{2}=-2$. Using point $A(-1,4)$ for $c$: $$4=-2(-1)+c \implies 4=2 + c \implies c=2.$$ Equation is $g(x)=-2x + 2$. 1.1.4 Write down range of $f(x)$: Since $a=-\frac{1}{2}<0$, parabola opens downward; max value at turning point $y=-3.5$. Actually calculate again: vertex $x=-\frac{b}{2a}=$ $$-\frac{-2}{2(-\frac{1}{2})} = -\frac{-2}{-1} = 2.$$ Calculate $f(2)$: $$f(2) = -\frac{1}{2}(2)^2 - 2(2) + \frac{5}{2} = -2 -4 + 2.5 = -3.5,$$ which is minimum; parabola opens downward but above vertex is max. Reconsider sign: Because $a=-1/2<0$, the parabola opens downward, vertex at maximum. So range: $$ (-\infty, 4] ext{ is incorrect; correct is } (-\infty, ) ?$$ But at $x= -1$, $f(-1)=4$ per data; vertex value is $-3.5$, so vertex is minimum or maximum? Since $a<0$, max value at vertex $f(2) = -3.5$. Thus range: $$(-\infty, 4]$$ does not fit; max value is $f$ at vertex $= -3.5$? No, this contradicts point A. Plot points $B, C$ at zeros, $A(-1,4)$ above vertex at $(2,-3.5)$. Hence parabola opens downward with max at vertex $(2, -3.5)$? But $A(-1,4)$ with $y=4$ higher. Contradiction, so re-check vertex calculation or leading coefficient sign: Vertex $x = -rac{b}{2a} = -rac{-2}{2 imes -1/2} = -rac{-2}{-1} = 2$ correct. $f(2) = -rac{1}{2}(2)^2 - 2(2) + rac{5}{2} = -2 - 4 + 2.5 = -3.5$. $f(-1) = -rac{1}{2}(-1)^2 - 2(-1) + rac{5}{2} = -rac{1}{2} + 2 + 2.5 = 4$. So point A is above vertex. Because $a<0$, vertex is max value, but here vertex $y=-3.5$ is less than $4$. Contradiction. Therefore, parabola opens downward with vertex giving max value; $f(-1)=4$ is less than max. Hence range is: $$(-rac{1}{2}x^2 - 2x + rac{5}{2}) o (-rac{1}{2})(x - (-2))^2 + k$$ vertex form: Convert to vertex form: $$f(x)=-rac{1}{2}x^2 - 2x + rac{5}{2}$$ $$= -rac{1}{2}(x^2 + 4x) + rac{5}{2}$$ Complete square inside parentheses: $$x^2 + 4x + 4 - 4 = (x+2)^2 - 4$$ So $$f(x) = -rac{1}{2}((x+2)^2 - 4) + rac{5}{2} = -rac{1}{2}(x+2)^2 + 2 + rac{5}{2} = -rac{1}{2}(x+2)^2 + rac{9}{2}.$$ Vertex is $( -2, rac{9}{2} )$ and because leading coefficient is negative, max value is $ rac{9}{2} = 4.5.$ So range: $$f(x) \\leq rac{9}{2}$$ Thus range is: $$(-\\infty, rac{9}{2}]$$. 1.2 Given $f(x) = \frac{6x - 12}{x + 3}$, draw graph with asymptotes and intercepts: Vertical asymptote: $x = -3$ (denominator zero). Horizontal asymptote: leading coeff of numerator and denominator degree equal: $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \frac{6x - 12}{x + 3} = 6.$$ Intercepts: - $y$-intercept at $x=0$: $$f(0) = \frac{6(0) - 12}{0 + 3} = \frac{-12}{3} = -4.$$ - $x$-intercept, set numerator zero: $$6x - 12 = 0 \Rightarrow x=2.$$ 1.3 Given $h(x)=5 \, 2^{-x} - 2$, draw graph with asymptote and intercepts: Rewrite: $h(x) = 5 \times 2^{-x} - 2 = 5 \times \frac{1}{2^x} - 2$. Horizontal asymptote: as $x \to \infty$, $2^{-x} \to 0$, so $h(x) \to -2$. Intercepts: - $y$-intercept: $h(0) = 5 \times 1 - 2 = 3$. - $x$-intercept: solve $$5 \times 2^{-x} - 2 =0 \implies 5 \times 2^{-x} = 2 \implies 2^{-x} = \frac{2}{5}.$$ Take logs: $$-x \log 2 = \log \frac{2}{5} \implies x = - \frac{\log \frac{2}{5}}{\log 2}.$$ 2.1.1 Simplify: $$\sin(360^\circ - x) \tan(-x) + \cos(180^\circ - x)(\sin^2 x + \cos^2 x)$$ Use identities: $$\sin(360^\circ - x) = \sin x,$$ $$\tan(-x) = -\tan x,$$ $$\cos(180^\circ - x) = -\cos x,$$ $$\sin^2 x + \cos^2 x = 1.$$ Thus expression: $$\sin x \times (-\tan x) + (-\cos x) \times 1 = -\sin x \tan x - \cos x.$$ Replace $\tan x = \frac{\sin x}{\cos x}$: $$-\sin x \times \frac{\sin x}{\cos x} - \cos x = -\frac{\sin^2 x}{\cos x} - \cos x = -\frac{\sin^2 x + \cos^2 x}{\cos x} = -\frac{1}{\cos x} = -\sec x.$$ 2.1.2 Simplify: $$\cos 135^\circ + \sin 225^\circ + \tan 330^\circ \sin 135^\circ.$$ Use exact values: $$\cos 135^\circ = -\frac{\sqrt{2}}{2},$$ $$\sin 225^\circ = -\frac{\sqrt{2}}{2},$$ $$\tan 330^\circ = -\frac{\sqrt{3}}{3},$$ $$\sin 135^\circ = \frac{\sqrt{2}}{2}.$$ Calculate: $$-\frac{\sqrt{2}}{2} + (-\frac{\sqrt{2}}{2}) + (-\frac{\sqrt{3}}{3}) \times \frac{\sqrt{2}}{2} = -\sqrt{2} + -\frac{\sqrt{6}}{6} = -\left( \sqrt{2} + \frac{\sqrt{6}}{6} \right).$$ 2.2 Prove: $$\frac{1}{\sin(180^\circ - x) + 1} - \frac{1}{\cos(90^\circ - x) - 1} = \frac{2}{\cos^2 x}.$$ Use identities: $$\sin(180^\circ - x) = \sin x,$$ $$\cos(90^\circ - x) = \sin x.$$ Left hand side: $$\frac{1}{\sin x + 1} - \frac{1}{\sin x - 1} = \frac{(\sin x - 1) - (\sin x + 1)}{(\sin x + 1)(\sin x - 1)} = \frac{\sin x - 1 - \sin x - 1}{\sin^2 x - 1} = \frac{-2}{\sin^2 x - 1}.$$ But $\sin^2 x -1 = -\cos^2 x$, so: $$= \frac{-2}{-\cos^2 x} = \frac{2}{\cos^2 x},$$ which is RHS. Thus proved. 2.3.1 Solve: $$3 \sin^2 x + 4 \sin x = 5.$$ Rewrite: $$3 s^2 + 4 s - 5 = 0, \quad s=\sin x.$$ Solve quadratic: $$s = \frac{-4 \pm \sqrt{16 + 60}}{6} = \frac{-4 \pm 8}{6}.$$ Solutions: $$s= \frac{4}{6} = \frac{2}{3}, \quad s= -2 ext{ (discard, not possible)}.$$ So $\sin x = \frac{2}{3}.$ Solutions: $$x = \sin^{-1}\left(\frac{2}{3}\right) + 360^\circ n, \quad 180^\circ - \sin^{-1}\left(\frac{2}{3}\right) + 360^\circ n.$$ 2.3.2 Solve: $$2 \sin x = \cos x.$$ Divide both sides by $\cos x$ (where $\cos x \neq 0$): $$2 \tan x = 1 \Rightarrow \tan x = \frac{1}{2}.$$ Solutions: $$x = \tan^{-1}\left(\frac{1}{2}\right) + 180^\circ n.$$ 2.4 Quadrilateral ABCD, Given: $$BC=120, AB=70,$$ Angles: $$\angle ADB=37.8^\circ, \angle DAB=82.4^\circ, \angle DBC=59.4^\circ.$$ 2.4.1 Find $BD$: In triangle $ADB$, use Law of Sines: $$\frac{BD}{\sin 82.4^\circ} = \frac{AB}{\sin 37.8^\circ} = \frac{70}{\sin 37.8^\circ}.$$ Thus: $$BD = \frac{70 \sin 82.4^\circ}{\sin 37.8^\circ} \approx \frac{70 \times 0.990}{0.613} \approx 112.9.$$ 2.4.2 Find $CD$: In triangle $BDC$, use Law of Cosines: $$CD^2 = BC^2 + BD^2 - 2 \times BC \times BD \times \cos(59.4^\circ).$$ Calculate: $$CD^2 = 120^2 + 112.9^2 - 2 \times 120 \times 112.9 \times \cos 59.4^\circ = 14400 + 12746.4 - 2 \times 120 \times 112.9 \times 0.507$$ $$= 27146.4 - 13733.6 = 13412.8,$$ $$CD = \sqrt{13412.8} \approx 115.75.$$ 2.4.3 Area of triangle BCD: Area formula: $$Area = \frac{1}{2} BC \times BD \times \sin(59.4^\circ)$$ $$= \frac{1}{2} \times 120 \times 112.9 \times 0.857 = 5814.3.$$ 2.4.4 Area of triangle ABD: Use given sides $AB=70$, $BD=112.9$, angle $\angle ADB=37.8^\circ$: $$Area = \frac{1}{2} AB \times BD \times \sin(37.8^\circ)$$ $$= \frac{1}{2} \times 70 \times 112.9 \times 0.613 = 2420.4.$$