I have opened the image you uploaded and will now begin solving the problems provided based on the diagram and questions shown. Let's start with the working and solutions for each part.

5.1: Determine the x-coordinates of A and B.

We are given the function $f(x) = -x^2 + 12$. This is a quadratic equation representing a parabola. The x-intercepts (points A and B) occur when $f(x) = 0$.

Set $f(x) = 0$: $-x^2 + 12 = 0$ $x^2 = 12$ $x = \pm \sqrt{12} = \pm 2\sqrt{3}$

So, the x-coordinates of A and B are $x = -2\sqrt{3}$ and $x = 2\sqrt{3}$.

5.2: Determine the range of $f$.

The function $f(x) = -x^2 + 12$ is a parabola that opens downwards (since the coefficient of $x^2$ is negative). The maximum value of $f(x)$ occurs at the vertex, which is at $x = 0$.

At $x = 0$, $f(0) = -(0)^2 + 12 = 12$

Therefore, the range of $f$ is $f(x) \in (-\infty, 12]$.

5.3: Determine the maximum value of $g(x) = 9 - 0.25x^2$.

The function $g(x) = 9 - 0.25x^2$ is also a parabola that opens downwards. The maximum value occurs at $x = 0$.

At $x = 0$, $g(0) = 9 - 0.25(0)^2 = 9$

Thus, the maximum value of $g(x)$ is 9.

5.4: Determine for which values of $x$, $f(x) = g(x)$.

We set $f(x) = g(x)$ and solve for $x$.

$-x^2 + 12 = 9 - 0.25x^2$ Multiply everything by 4 to eliminate the decimal: $-4x^2 + 48 = 36 - x^2$ Move all terms to one side: $-4x^2 + x^2 = 36 - 48$ $-3x^2 = -12$ $x^2 = 4$ $x = \pm 2$

So, $f(x) = g(x)$ when $x = -2$ and $x = 2$.

5.5: For which real values of $x$ will $f(x) > g(x)$?

We now need to find where $f(x) > g(x)$. From the previous solution, we know the points where $f(x) = g(x)$ are $x = -2$ and $x = 2$.

• When $x \in (-2, 2)$, $f(x) > g(x)$, because between these points, $f(x)$ is greater than $g(x)$, as the curves are crossing each other.

Thus, $f(x) > g(x)$ for $-2 < x < 2$.

5.6: Write down the equation of the line passing through points A and T.

Point A is $(-2\sqrt{3}, 0)$ and point T is $(0, 12)$.

The slope of the line is: $\text{slope} = \frac{12 - 0}{0 - (-2\sqrt{3})} = \frac{12}{2\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$

The equation of the line is in the form $y = mx + c$. We know the slope $m = 2\sqrt{3}$, and the y-intercept $c = 12$, since it passes through the point $(0, 12)$.

Thus, the equation of the line is: $y = 2\sqrt{3}x + 12$

These are the working out and solutions for the problems in the image you provided. Would you like any further details on these solutions or another explanation?

Here are five related questions to extend your learning:

1. How do you find the vertex of a parabola given its equation?
2. What is the significance of the discriminant in solving quadratic equations?
3. How would you find the points of intersection between two parabolas analytically?
4. What methods can you use to determine whether a quadratic opens upwards or downwards?
5. Can you describe the graphical behavior of a quadratic equation when the leading coefficient is negative?

Tip: When solving quadratic inequalities, graphing the functions helps visualize where one function is greater than the other.