1.1.1 State the problem: Show that the equation of $f(x)$ is $-\frac{1}{2}x^2 - 2x + \frac{5}{2}$. 1. Use points B(-5,0) and C(1,0) where $f(x)=0$ to establish equations. 2. For $f(x)=ax^2+bx+c$, B gives $0=a(-5)^2 + b(-5)+c=25a - 5b + c$ and C gives $0=a(1)^2 + b(1)+c = a + b + c$. 3. Since A(-1,4) lies on both $f$ and $g$, substitute into $f$: $4 = a(-1)^2 + b(-1) + c = a - b + c$. 4. Solve system: $$\begin{cases} 25a -5b + c = 0 \\ a + b + c = 0 \\ a - b + c = 4 \end{cases}$$ 5. Subtract second from third equation: $(a - b + c) - (a + b + c) = 4 - 0 \Rightarrow -2b =4 \Rightarrow b = -2$. 6. Use $a + b + c=0 \Rightarrow c = -a - b = -a +2$. 7. Substitute $b=-2$ and $c=-a + 2$ into first equation: $$25a -5(-2) + (-a + 2) =0 \Rightarrow 25a + 10 - a + 2=0 \Rightarrow 24a + 12=0 \Rightarrow a=-\frac{12}{24}=-\frac{1}{2}.$$ 8. Substitute $a$ back to find $c$: $$c = -a + 2 = -\left(-\frac{1}{2}\right) + 2 = \frac{1}{2} + 2 = \frac{5}{2}.$$ 9. Therefore $f(x) = -\frac{1}{2}x^2 - 2x + \frac{5}{2}$. 1.1.2 Determine turning point of $f(x)$. 1. For $f(x)=ax^2 + bx + c$ the vertex $x$-coordinate is $x = -\frac{b}{2a}$. 2. Here, $a = -\frac{1}{2}$ and $b = -2$. 3. Calculate: $$x = -\frac{-2}{2(-1/2)} = -\frac{-2}{-1} = -2.$$ 4. Calculate $y$ at $x=-2$: $$f(-2) = -\frac{1}{2}(-2)^2 - 2(-2) + \frac{5}{2} = -\frac{1}{2} \times 4 + 4 + \frac{5}{2} = -2 + 4 + 2.5 = 4.5.$$ 5. Turning point is $(-2, 4.5)$. 1.1.3 Determine equation of $g(x)=mx + c$. 1. Points of $g$ are given: it passes through A(-1,4) and C(1,0). 2. Gradient: $$m = \frac{0 - 4}{1 - (-1)} = \frac{-4}{2} = -2.$$ 3. Use point-slope form with point A: $$4 = -2(-1) + c \Rightarrow 4 = 2 + c \Rightarrow c = 2.$$ 4. Equation: $g(x) = -2x + 2$. 1.1.4 Write down range of $f(x)$. 1. Since $f$ is a downward parabola (because $a = -\frac{1}{2} < 0$), maximum value is at turning point. 2. The maximum value is $4.5$ at $x=-2$. 3. Range: $(-\infty, 4.5]$. 1.2 Draw and analyze $f(x) = \frac{6x - 12}{x + 3}$. 1. Find vertical asymptote by setting denominator zero: $$x + 3 = 0 \Rightarrow x = -3.$$ 2. Find horizontal asymptote: Degree numerator and denominator both 1. Leading coefficients ratio: $$y = \frac{6x}{x} = 6.$$ Horizontal asymptote is $y=6$. 3. Find $x$-intercept: Set numerator zero: $$6x - 12=0 \Rightarrow x=2.$$ $x$-intercept: $(2,0)$. 4. Find $y$-intercept: Set $x=0$: $$f(0) = \frac{6(0)-12}{0+3} = \frac{-12}{3} = -4.$$ $y$-intercept: $(0, -4)$. 4. Graph has vertical asymptote at $x=-3$, horizontal asymptote at $y=6$, passes through points $(2,0)$ and $(0,-4)$. 1.3 For $h(x) = 5.2^{-x} - 2$. 1. Identify horizontal asymptote: As $x \to \infty$, $2^{-x} \to 0$, so $$h(x) \to -2.$$ Horizontal asymptote: $y = -2$. 2. Find $y$-intercept: $x=0$ $$h(0) = 5.2^{0} - 2 = 5 - 2 =3.$$ 3. Find $x$-intercept: Set $h(x) = 0$: $$0 = 5.2^{-x} - 2 \Rightarrow 5.2^{-x} = 2 \Rightarrow 2^{-x} = \frac{2}{5}.$$ Take logarithm base 2: $$-x = \log_2 \left( \frac{2}{5} \right) = \log_2 2 - \log_2 5 = 1 - \log_2 5.$$ So: $$x = \log_2 5 - 1.$$ Approximate: $$\log_2 5 \approx 2.322, \quad x \approx 2.322 -1 = 1.322.$$ $x$-intercept approximately $(1.32, 0)$. 2.1.1 Simplify: $$\sin(360^{\circ} - x) \tan(-x) + \cos(180^{\circ} - x)(\sin^2 x + \cos^2 x).$$ 1. Use identities: $$\sin(360^{\circ} - x) = -\sin x, \quad \tan(-x) = -\tan x, \quad \cos(180^{\circ} - x) = -\cos x,$$ $$\sin^2 x + \cos^2 x = 1.$$ 2. Substitute: $$(-\sin x)(-\tan x) + (-\cos x)(1) = \sin x \tan x - \cos x.$$ 3. Since $\tan x = \frac{\sin x}{\cos x}$: $$\sin x \times \frac{\sin x}{\cos x} - \cos x = \frac{\sin^2 x}{\cos x} - \cos x = \frac{\sin^2 x - \cos^2 x}{\cos x}.$$ 4. Use identity $\sin^2 x - \cos^2 x = -\cos 2x$. 5. Final: $$\frac{-\cos 2x}{\cos x} = -\frac{\cos 2x}{\cos x}.$$ 2.1.2 Simplify: $$\cos 135^{\circ} + \sin 225^{\circ} + \tan 330^{\circ} \sin 135^{\circ}.$$ 1. Evaluate each: $$\cos 135^{\circ} = -\frac{\sqrt{2}}{2}, \quad \sin 225^{\circ} = -\frac{\sqrt{2}}{2},$$ $$\tan 330^{\circ} = -\frac{\sqrt{3}}{3}, \quad \sin 135^{\circ} = \frac{\sqrt{2}}{2}.$$ 2. Substitute: $$-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + \left(-\frac{\sqrt{3}}{3}\right) \times \frac{\sqrt{2}}{2} = -\sqrt{2} - \frac{\sqrt{6}}{6}.$$ 2.2 Prove: $$\frac{1}{\sin(180^{\circ} - x) + 1} - \frac{1}{\cos(90^{\circ} - x) - 1} = \frac{2}{\cos^2 x}.$$ 1. Use identities: $$\sin(180^{\circ} - x) = \sin x, \quad \cos(90^{\circ} - x) = \sin x.$$ 2. Left side becomes: $$\frac{1}{\sin x + 1} - \frac{1}{\sin x - 1} = \frac{(\sin x - 1) - (\sin x + 1)}{(\sin x + 1)(\sin x - 1)} = \frac{\sin x - 1 - \sin x - 1}{\sin^2 x - 1} = \frac{-2}{\sin^2 x - 1}.$$ 3. Since $\sin^2 x + \cos^2 x = 1$, then $\sin^2 x - 1 = -\cos^2 x$. 4. Substitute: $$\frac{-2}{-\cos^2 x} = \frac{2}{\cos^2 x}.$$ 5. Proved. 2.3.1 Solve $3 \sin^2 x + 4 \sin x = 5$. 1. Rearrange: $$3\sin^2 x + 4 \sin x - 5 = 0.$$ 2. Substitute $u = \sin x$: $$3u^2 + 4u -5 = 0.$$ 3. Solve quadratic: $$u = \frac{-4 \pm \sqrt{16 + 60}}{6} = \frac{-4 \pm \sqrt{76}}{6} = \frac{-4 \pm 2\sqrt{19}}{6} = \frac{-2 \pm \sqrt{19}}{3}.$$ 4. Approximate roots: $$u_1 = \frac{-2 + 4.359}{3} = 0.786, \quad u_2 = \frac{-2 - 4.359}{3} = -2.12 ext{ (invalid as } \sin x \in [-1,1])$$ 5. Valid: $\sin x = 0.786$. 6. Solutions on $[0,360^{\circ})$: $$x = \sin^{-1}(0.786) = 51.9^{\circ}, \quad 180^{\circ} - 51.9^{\circ} = 128.1^{\circ}.$$ 2.3.2 Solve $2 \sin x = \cos x$. 1. Rearrange: $$\frac{\sin x}{\cos x} = \frac{1}{2} \Rightarrow \tan x = \frac{1}{2}.$$ 2. Find general solutions: $$x = \tan^{-1}(1/2) + n\pi = 26.57^{\circ} + n180^{\circ}, n \in \mathbb{Z}.$$ 3. On $[0, 360^{\circ})$: $$x = 26.57^{\circ}, 206.57^{\circ}.$$ 2.4 Quadrilateral ABCD analysis. Given: $$BC = 120, AB=70,$$ $$\angle ADB =37.8^{\circ}, \angle DAB=82.4^{\circ}, \angle DBC=59.4^{\circ}.$$ 2.4.1 Find length BD. 1. Use Law of Sines on triangle ABD or apply geometric relations (requires more info - assume triangle ABD and use Law of Cosines). 2. Given angles and sides, use Law of Cosines: $$BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle DAB).$$ 3. Need length AD. Using triangle with angles and sides we can find AD using Law of Sines. 4. Approach: In triangle ABD, $$\angle ADB = 37.8^{\circ}, \angle DAB = 82.4^{\circ}.$$ Calculate $\angle ABD$: $$\angle ABD = 180^{\circ} - 82.4^{\circ} - 37.8^{\circ} = 59.8^{\circ}.$$ 5. Using Law of Sines to find AD: $$\frac{AD}{\sin 59.8^{\circ}} = \frac{AB}{\sin 37.8^{\circ}} \Rightarrow AD = \frac{70 \times \sin 59.8^{\circ}}{\sin 37.8^{\circ}}.$$ 6. Approximate: $$\sin 59.8^{\circ} \approx 0.865, \sin 37.8^{\circ} \approx 0.613.$$ $$AD = \frac{70 \times 0.865}{0.613} \approx 98.8.$$ 7. Find BD using Law of Cosines: $$BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle ADB) = 70^2 + 98.8^2 - 2 \times 70 \times 98.8 \times \cos 37.8^{\circ}.$$ 8. Calculate: $$70^2 = 4900, \quad 98.8^2 \approx 9764. $$ $$\cos 37.8^{\circ} \approx 0.79.$$ $$BD^2 = 4900 + 9764 - 2 \times 70 \times 98.8 \times 0.79 = 14664 - 10928 \approx 3736.$$ 9. So: $$BD = \sqrt{3736} \approx 61.14.$$ 2.4.2 Determine length CD. 1. Use Law of Sines on triangle BCD. Known: $$BC = 120, \angle DBC = 59.4^{\circ}, \angle BDC = \angle ADB = 37.8^{\circ}$$ (assuming quadrilateral angles, angle at D in BCD same as ADB). Calculate $\angle BCD$: $$180^{\circ} - 59.4^{\circ} - 37.8^{\circ} = 82.8^{\circ}.$$ 2. Apply Law of Sines: $$\frac{CD}{\sin 59.4^{\circ}} = \frac{BC}{\sin 82.8^{\circ}} \Rightarrow CD = \frac{120 \times \sin 59.4^{\circ}}{\sin 82.8^{\circ}}.$$ 3. Approximate: $$\sin 59.4^{\circ} \approx 0.86, \sin 82.8^{\circ} \approx 0.99.$$ $$CD = \frac{120 \times 0.86}{0.99} \approx 104.2.$$ 2.4.3 Determine area of triangle BCD. 1. Use formula area = $\tfrac{1}{2}ab \sin C$ for sides $BC$, $CD$ and included angle $\angle BCD$. 2. Known: $$BC = 120, CD = 104.2, \angle BCD = 82.8^{\circ}.$$ 3. Compute area: $$= \frac{1}{2} \times 120 \times 104.2 \times \sin 82.8^{\circ}.$$ 4. Approximate $\sin 82.8^{\circ} \approx 0.99$: $$\approx 0.5 \times 120 \times 104.2 \times 0.99 = 6190.$$ 2.4.4 Determine area of triangle ABD. 1. Use formula area = $\tfrac{1}{2}ab \sin C$ using sides $AB=70$, $AD=98.8$, and included angle $\angle ADB = 37.8^{\circ}$. 2. Compute: $$= \frac{1}{2} \times 70 \times 98.8 \times \sin 37.8^{\circ}.$$ 3. Approximate $\sin 37.8^{\circ} \approx 0.613$. $$= 0.5 \times 70 \times 98.8 \times 0.613 = 2118.$$ Final answers: $f(x) = -\frac{1}{2}x^2 - 2x + \frac{5}{2}$. Turning point: $(-2,4.5)$. $g(x) = -2x + 2$. Range of $f(x)$: $(-\infty, 4.5]$. $f(x) = \frac{6x - 12}{x + 3}$ has vertical asymptote $x=-3$, horizontal $y=6$, intercepts $(2,0)$ and $(0,-4)$. $h(x) = 5.2^{-x} - 2$ has horizontal asymptote $y=-2$, intercepts $(0,3)$ and approximately $(1.32, 0)$. $2.1.1$ simplifies to $-\frac{\cos 2x}{\cos x}$. $2.1.2$ simplifies to $-\sqrt{2} - \frac{\sqrt{6}}{6}$. $2.2$ proven true. Solutions for $2.3.1$: $51.9^{\circ}, 128.1^{\circ}$. For $2.3.2$: $26.57^{\circ}, 206.57^{\circ}$. Lengths in $2.4$: $BD \approx 61.14$, $CD \approx 104.2$. Areas $BCD \approx 6190$, $ABD \approx 2118$.