**1. Stating the problem:** We are given a quadratic function $$y=2x^2 - 3$$ and a partially filled table of values for $$x = -3, -2, -1, 0, 1, 2, 3$$ with corresponding $$y$$ values as $$15, 5, -1, -3, ???, 5, 15$$. We need to: (i) Find the value of $$y$$ when $$x=1$$. (ii) Draw the graph of the function using the given scale. (iii) Write the minimum value of the function. (iv) Find the interval of $$x$$ values for which the function decreases or increases. (v) Find the roots of the equation $$2x^2 -3 =0$$. **2. Find the missing value $$y$$ for $$x=1$$:** Substitute $$x=1$$ into the function: $$y=2(1)^2 -3 = 2(1) - 3 = 2 - 3 = -1$$ So the missing value is $$-1$$. **3. Graphing the function:** - The function is $$y=2x^2 -3$$ which is a parabola opening upwards. - Table of values now: | x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | | y | 15 | 5 | -1 | -3 | -1 | 5 | 15 | - Scale for x-axis: 10 small divisions = 1 unit. - Scale for y-axis: 10 small divisions = 2 units. - Plot the points and draw a smooth curve through them to form the parabola with vertex at (0, -3). **4. Minimum value of the function:** - The parabola opens upwards. - The vertex is the minimum point. - Vertex at $$x=0$$, $$y=-3$$. Minimum value of $$y$$ is $$-3$$ at $$x=0$$. **5. Intervals where the function increases or decreases:** - Because the parabola opens upwards and is symmetric: - It decreases on $$(-\infty, 0)$$. - It increases on $$(0, +\infty)$$. **6. Roots of the equation $$2x^2 - 3 = 0$$:** Solve for $$x$$: $$ 2x^2 - 3 = 0 \Rightarrow 2x^2 = 3 \Rightarrow x^2 = \frac{3}{2} \Rightarrow x = \pm \sqrt{\frac{3}{2}} = \pm \frac{\sqrt{6}}{2} $$ **Final answers:** (i) $$y = -1$$ at $$x=1$$. (ii) The graph is a parabola opening upwards with vertex at $$(0, -3)$$. (iii) Minimum value is $$-3$$ at $$x=0$$. (iv) Decreasing on $$(-\infty, 0)$$, increasing on $$(0, +\infty)$$. (v) Roots are $$x = \pm \frac{\sqrt{6}}{2}$$.