The problem provides the graph of a function of the form $y = b - ax^2$. It asks to find the maximum value of the function, the equation of the function, and the range of $x$ values for which the function is positive. Then, the problem asks to find the equation of the graph when the $y$ values are minimized while the $x$-intercepts stay the same, and find the axis of symmetry and the turning point coordinates. Finally, there is a separate simple interest problem to be solved, but since the graph related questions are the focus, let's concentrate on these questions.

1. Problem Description

The problem provides the graph of a function of the form $y = b - ax^2$. It asks to find the maximum value of the function, the equation of the function, and the range of $x$ values for which the function is positive. Then, the problem asks to find the equation of the graph when the $y$ values are minimized while the $x$-intercepts stay the same, and find the axis of symmetry and the turning point coordinates. Finally, there is a separate simple interest problem to be solved, but since the graph related questions are the focus, let's concentrate on these questions.

2. Solution Steps

a) i) The maximum value of the function is the $y$-coordinate of the vertex. From the graph, the vertex is at $(0, 6)$. So the maximum value is

6.

a) ii) The function is of the form $y = b - ax^2$. The vertex is at $(0, b)$, so $b = 6$. We need to find $a$. The graph passes through the point $(1, 2)$. Plugging this into the equation:

$2 = 6 - a(1)^2$

$2 = 6 - a$

$a = 6 - 2 = 4$

So the equation of the function is $y = 6 - 4x^2$.

a) iii) We need to find when $y > 0$. So we need to solve the inequality:

$6 - 4x^2 > 0$

$6 > 4x^2$

$x^2 < \frac{6}{4} = \frac{3}{2}$

$-\sqrt{\frac{3}{2}} < x < \sqrt{\frac{3}{2}}$

Approximately, $-1.22 < x < 1.22$. From the graph, $x$ is between $-1.2$ and $1.2$ (approximately). So the range of x values is $(-1.2, 1.2)$.

b) i) The roots stay the same when the graph is minimum, so $y=-6+4x^2$ is a reflection about the x-axis. So equation is $y = 4x^2 -6$.

b) ii) The axis of symmetry is the vertical line passing through the vertex. The vertex is $(0, -6)$. So, the axis of symmetry is $x = 0$. The turning point is the vertex, so the coordinates are $(0, -6)$.

3. Final Answer

a) i) 6

a) ii) $y = 6 - 4x^2$

a) iii) $(- \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$ or approximately $(-1.22, 1.22)$

b) i) $y = 4x^2 - 6$

b) ii) Axis of symmetry: $x = 0$, Turning point: $(0, -6)$