The sketch below shows the graphs of $f(x) = x^2 - 2x - 3$ and $g(x) = x - 3$ - NSC Mathematics - Question 5 - 2017 - Paper 1

The coordinates of A are $(-1, 0)$ and B are $(3, 0)$.

To calculate the length of AB, we find:

Length = $|3 - (-1)| = |3 + 1| = 4$ units.

The turning point D of the parabola $f(x)$ occurs at the vertex, which can be found using the formula $x = -\frac{b}{2a}$.

Here, $a = 1$ and $b = -2$, so:

$x = -\frac{-2}{2 \cdot 1} = 1$.

Substituting back into $f$ to find the y-coordinate:

$y = f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$.

Therefore, the coordinates of D are $(1, -4)$.

The average gradient between two points can be calculated by:

$\text{Gradient} = \frac{f(D) - f(C)}{x_D - x_C}$

Where:

• $f(C) = 0$ (the y-coordinate of C)
• $f(D) = -4$ (the y-coordinate of D)
• $x_C = 3$
• $x_D = 1$

Thus:

Gradient = $\frac{-4 - 0}{1 - 3} = \frac{-4}{-2} = 2$.

To find the size of $riangle OCB$, we note that O is at the origin (0,0), B is at (3,0), and C is at (3,0).

Using the coordinates to form a right-angled triangle, the lengths OC and OB are equal (both are 3 units).

Thus, we can conclude that:

$riangle OCB$ is isosceles, and the angle at C can be calculated using the standard properties of angles in triangles.

For $f(x) = k$ to have two unequal positive real roots, the discriminant must be positive. The equation can be rewritten as:

$x^2 - 2x - (3 + k) = 0$

The discriminant $D$ is given by:

$D = b^2 - 4ac = (-2)^2 - 4(1)(-3 - k) = 4 + 12 + 4k = 16 + 4k$

For the roots to be unequal and positive, we set $D > 0$ and check the sign:

$16 + 4k > 0 \Rightarrow k > -4$.

The second derivative of $f(x)$ is constant since:

$f(x) = x^2 - 2x - 3$ $f''(x) = 2$

Since $f''(x) = 2 > 0$ for all x, thus $f''(x) > 0$ for all values of $x$.