1. **State the problem:** We are given the quadratic function $$f(x) = 2x^2 - 9x - 5$$ and asked to find: (i) The coordinates of point C where the function crosses the y-axis. (ii) The coordinates of points A and B where the function crosses the x-axis. (iii) The slope of the tangent to the function at $$x=4$$. 2. **Recall important formulas:** - The y-intercept occurs where $$x=0$$, so $$f(0)$$ gives the y-coordinate. - The x-intercepts occur where $$f(x) = 0$$, so solve $$2x^2 - 9x - 5 = 0$$. - The slope of the tangent at a point is given by the derivative $$f'(x)$$ evaluated at that point. 3. **Find the y-intercept (point C):** $$f(0) = 2(0)^2 - 9(0) - 5 = -5$$ So, point C is at $$(0, -5)$$. 4. **Find the x-intercepts (points A and B):** Solve $$2x^2 - 9x - 5 = 0$$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=-9$$, $$c=-5$$. Calculate the discriminant: $$\Delta = (-9)^2 - 4(2)(-5) = 81 + 40 = 121$$ Calculate the roots: $$x = \frac{9 \pm \sqrt{121}}{2 \times 2} = \frac{9 \pm 11}{4}$$ So, $$x_1 = \frac{9 + 11}{4} = \frac{20}{4} = 5$$ $$x_2 = \frac{9 - 11}{4} = \frac{-2}{4} = -0.5$$ Points A and B are at $$(5, 0)$$ and $$(-0.5, 0)$$ respectively. 5. **Find the slope of the tangent at $$x=4$$:** First, find the derivative: $$f'(x) = \frac{d}{dx}(2x^2 - 9x - 5) = 4x - 9$$ Evaluate at $$x=4$$: $$f'(4) = 4(4) - 9 = 16 - 9 = 7$$ So, the slope of the tangent at $$x=4$$ is $$7$$. **Final answers:** - Point C: $$(0, -5)$$ - Points A and B: $$(5, 0)$$ and $$(-0.5, 0)$$ - Slope at $$x=4$$: $$7$$