The sketch shows the graph of $f(x) = x(x + 3)$ and $g(x) = -\frac{1}{2}x + 2$ - NSC Mathematics - Question 5 - 2017 - Paper 1

To find the turning point of $f$, we take the derivative $f'(x)$ and set it to zero: $f'(x) = 2x + 3$

Setting the derivative equal to zero: $2x + 3 = 0$ $x = -\frac{3}{2}$

Now we substitute $x$ back into the function to find the y-coordinate: $f(-\frac{3}{2}) = -\frac{3}{2}(-\frac{3}{2} + 3) = -\frac{3}{2}(\frac{3}{2}) = -\frac{9}{4}$

Thus, the coordinates of point P are: $P(-\frac{3}{2}; -\frac{9}{4})$

To find the average gradient between points $x = -5$ and $x = -3$, we calculate: $f(-5) = -5(-5 + 3) = 10$ $f(-3) = -3(-3 + 3) = 0$

Average gradient formula: $m = \frac{f(-3) - f(-5)}{-3 - (-5)} = \frac{0 - 10}{-3 + 5} = \frac{-10}{2} = -5$

To find where $f(x) > 0$, we need to look at the roots found earlier: $f(x) = x(x + 3)$ The roots are $x = 0$ and $x = -3$.

The function will be positive between the roots. Testing intervals:

• For $x < -3$, $f(x) < 0$
• For $-3 < x < 0$, $f(x) > 0$
• For $x > 0$, $f(x) > 0$

Thus, $f(x) > 0$ for: $x \in (-3, 0) \cup (0, \infty)$

First, we find the turning point of $f(x)$, which we determined to be at $x = -\frac{3}{2}$. For $h(x)$, we apply the transformation: $h(x) = f(x - 2)$

The new turning point occurs at: $x = -\frac{3}{2} + 2 = \frac{1}{2}$

We then calculate the y-coordinate: $h(\frac{1}{2}) = f(\frac{1}{2} - 2) = f(-\frac{3}{2}) = -\frac{9}{4}$

Thus, the turning point of $h$ is: $h(\frac{1}{2}; -\frac{9}{4})$

Given the equations of the straight line and the parabola: $g(x) = -\frac{1}{2}x + 2$ $h(x) = f(x-2) = -\frac{1}{2}x + 2$

Using the distance formula for $LM$, where: $LM = d = \sqrt{(x_L - x_M)^2 + (y_L - y_M)^2}$ Substituting the respective coordinates, we can simplify: $LM = \left(-\frac{x + 7}{4}\right)^2 \cdot \frac{81}{16}$ which can be derived using the expressions for the lengths along the corresponding axes.