Question: Worksheet 7 V (Test) 6. The equation of the line through the points A and C are given by y = 3x + 9 6.1 Determine the equation of the parabola. 6.2 Determine the equation of the straight line h if 3OD = OC 6.3 Determine the coordinates of Point F, the intersection of h and g. 6.4 Determine the coordinates of Point E, the intersection of h and f. 6.5 Determine the average gradient of the parabola between the Points C and E. 6.6 What is the domain and range of the parabola? 7. Given: f(x) = -4/x + 1 g(x) = ab^x The graphs intersect at D(-1;5) 7.1 Determine the equation of g(x). 7.2 Determine the domain of f. 7.3 Give the equations of the asymptotes of f. 7.4 Give the reflection of g in the y-axis. 7.5 Determine the equations of the axes of symmetry of f. 7.6 Determine the length of AB if OC = 2 units. 8. Draw rough sketches of: y = ax^2 + q if 8.1 a > 0 and q < 0 8.2 a < 0 and q = 0 8.3 a = 0 and q > 0 9. Draw rough sketches of: y = a/x + q if 9.1 a > 0 and q = 0 9.2 a < 0 and q > 0 9.3 a = 0 and q < 0 Graph positions: Top-left: a parabola opening downward, vertex at C on the y-axis, crossing x-axis at A and B; line g rises through A and C; line h slopes downward through E, D, and B; F is the intersection of g and h. Bottom-left: f(x) = -4/x + 1 is a hyperbola with asymptotes x = 0 and y = 1, and g(x) = ab^x is an increasing exponential passing through D(-1;5), E on the y-axis, and A/B on the x-axis.

1. **Problem Statement:** We are given the line through points A and C as $$y = 3x + 9$$ and several questions about the parabola, lines, intersections, gradients, and domains. 2. **Determine the equation of the parabola (6.1):** - The parabola opens downward with vertex at C on the y-axis. - Assume the parabola is of the form $$y = ax^2 + bx + c$$. - Since vertex C is on the y-axis, its x-coordinate is 0, so vertex at (0, c). - Given the parabola opens downward, $$a < 0$$. - The line through A and C is $$y = 3x + 9$$, so point C is on this line at $$x=0$$, giving $$y=9$$. - So vertex C is at (0, 9), so $$c = 9$$. 3. **Find points A and B (x-intercepts):** - The parabola crosses the x-axis at A and B, so $$y=0$$. - Set $$0 = ax^2 + bx + 9$$. 4. **Use point A on line g:** - Since line g passes through A and C, and equation is $$y=3x+9$$. - At A, $$y=0$$, so $$0=3x_A + 9 \\ \Rightarrow x_A = -3$$. - So point A is $$(-3, 0)$$. 5. **Use point A in parabola equation:** - $$0 = a(-3)^2 + b(-3) + 9 = 9a - 3b + 9$$. 6. **Use point B (other x-intercept):** - Let $$x_B = m$$, then $$0 = a m^2 + b m + 9$$. 7. **Use vertex form to relate a and b:** - Vertex at $$x=0$$ means $$b = 0$$ (since vertex $$x = -\frac{b}{2a}$$). - So $$b=0$$. 8. **Simplify with $$b=0$$:** - From step 5: $$0 = 9a + 9 \\ \Rightarrow 9a = -9 \\ \Rightarrow a = -1$$. 9. **Equation of parabola:** - $$y = -x^2 + 9$$. 10. **Equation of line h if $$3OD = OC$$ (6.2):** - Let points O, D, C be on a line with coordinates such that $$3OD = OC$$. - If $$OD = d$$, then $$OC = 3d$$. - Since C is at (0,9), O at origin (0,0), then $$OC = 9$$ units. - So $$OD = 3$$ units. - Point D is at (0,3). - Line h passes through points D and B. - We need coordinates of B. 11. **Find B (other x-intercept of parabola):** - Set $$y=0$$ in parabola: $$0 = -x^2 + 9 \\ \Rightarrow x^2 = 9 \\ \Rightarrow x = \pm 3$$. - A is at (-3,0), so B is at (3,0). 12. **Equation of line h through D(0,3) and B(3,0):** - Slope $$m = \frac{0 - 3}{3 - 0} = \frac{-3}{3} = -1$$. - Equation: $$y - 3 = -1(x - 0) \\ y = -x + 3$$. 13. **Coordinates of Point F, intersection of h and g (6.3):** - Line g: $$y = 3x + 9$$. - Line h: $$y = -x + 3$$. - Set equal: $$3x + 9 = -x + 3 \\ 3x + x = 3 - 9 \\ 4x = -6 \\ x = -\frac{3}{2}$$. - Substitute back: $$y = 3(-\frac{3}{2}) + 9 = -\frac{9}{2} + 9 = \frac{9}{2} = 4.5$$. - So $$F = (-1.5, 4.5)$$. 14. **Coordinates of Point E, intersection of h and f (6.4):** - Function $$f(x) = -\frac{4}{x} + 1$$. - Line h: $$y = -x + 3$$. - Set equal: $$-\frac{4}{x} + 1 = -x + 3 \\ -\frac{4}{x} + 1 + x - 3 = 0 \\ x - \frac{4}{x} - 2 = 0$$. - Multiply both sides by $$x$$ (assuming $$x \neq 0$$): $$x^2 - 4 - 2x = 0 \\ x^2 - 2x - 4 = 0$$. - Solve quadratic: $$x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}$$. - Two possible x-values: $$1 + \sqrt{5}$$ and $$1 - \sqrt{5}$$. - Calculate corresponding y-values: For $$x = 1 + \sqrt{5}$$: $$y = -x + 3 = - (1 + \sqrt{5}) + 3 = 2 - \sqrt{5}$$. For $$x = 1 - \sqrt{5}$$: $$y = - (1 - \sqrt{5}) + 3 = 2 + \sqrt{5}$$. - So points E are approximately: $$E_1 = (1 + \sqrt{5}, 2 - \sqrt{5})$$, $$E_2 = (1 - \sqrt{5}, 2 + \sqrt{5})$$. 15. **Average gradient of parabola between points C and E (6.5):** - Use point C (0,9) and point E (choose $$E_1$$ for example). - Coordinates: $$C = (0,9)$$, $$E = (1 + \sqrt{5}, 2 - \sqrt{5})$$. - Average gradient: $$m = \frac{y_E - y_C}{x_E - x_C} = \frac{(2 - \sqrt{5}) - 9}{(1 + \sqrt{5}) - 0} = \frac{-7 - \sqrt{5}}{1 + \sqrt{5}}$$. - Rationalize denominator: $$m = \frac{-7 - \sqrt{5}}{1 + \sqrt{5}} \times \frac{1 - \sqrt{5}}{1 - \sqrt{5}} = \frac{(-7 - \sqrt{5})(1 - \sqrt{5})}{1 - 5} = \frac{(-7)(1) + 7\sqrt{5} - \sqrt{5} + 5}{-4}$$ $$= \frac{-7 + 7\sqrt{5} - \sqrt{5} + 5}{-4} = \frac{-2 + 6\sqrt{5}}{-4} = \frac{-2}{-4} + \frac{6\sqrt{5}}{-4} = \frac{1}{2} - \frac{3\sqrt{5}}{2}$$. - So average gradient is $$\frac{1}{2} - \frac{3\sqrt{5}}{2}$$. 16. **Domain and range of the parabola (6.6):** - Domain: all real numbers $$\mathbb{R}$$. - Range: since parabola opens downward with vertex at 9, range is $$y \leq 9$$. **Final answers:** - Equation of parabola: $$y = -x^2 + 9$$. - Equation of line h: $$y = -x + 3$$. - Coordinates of F: $$(-1.5, 4.5)$$. - Coordinates of E: $$\left(1 \pm \sqrt{5}, 2 \mp \sqrt{5}\right)$$. - Average gradient between C and E: $$\frac{1}{2} - \frac{3\sqrt{5}}{2}$$. - Domain of parabola: $$\mathbb{R}$$. - Range of parabola: $$y \leq 9$$.