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 the function $f(x) = x^2 + 3x$, we first compute the vertex using the formula for x-coordinate: $x = -\frac{b}{2a} = -\frac{3}{2}$ Substituting $x = -\frac{3}{2}$ back into $f$ gives: $f(-\frac{3}{2}) = (-\frac{3}{2})^2 + 3(-\frac{3}{2}) = -\frac{9}{4}$ Thus, the coordinates of P are igg(-\frac{3}{2}; -\frac{9}{4}\bigg).

The average gradient of a function $f$ over an interval $[x_1, x_2]$ is computed as: $\frac{f(x_2) - f(x_1)}{x_2 - x_1}$ Here, $x_1 = -5$ and $x_2 = -3$. We first find $f(-5)$ and $f(-3)$:

• $f(-5) = (-5)(-5 + 3) = (-5)(-2) = 10$
• $f(-3) = (-3)(-3 + 3) = 0$ Calculating the average gradient gives: $\frac{0 - 10}{-3 - (-5)} = \frac{-10}{2} = -5$

To determine where $f(x) > 0$, we solve the inequality: $x(x + 3) > 0$ The critical points are found by setting $f(x) = 0$, which gives $x = 0$ and $x = -3$. Testing intervals:

• For $x < -3$, choose $x = -4$: $(-4)(-1) > 0 ightarrow$ True
• For $-3 < x < 0$, choose $x = -1$: $(-1)(2) < 0 ightarrow$ False
• For $x > 0$, choose $x = 1$: $(1)(4) > 0 ightarrow$ True Thus, $f(x) > 0$ for $x < -3$ and $x > 0$.

The function $h(x) = f(x - 2)$ represents a horizontal shift of $f$. Thus, the coordinates of the turning point of $h(x)$ can be found by substituting $x = -\frac{3}{2} + 2 = \frac{1}{2}$ into $h$: $h(\frac{1}{2}) = f(\frac{1}{2} - 2) = f(-\frac{3}{2}) = -\frac{9}{4}$ Therefore, the coordinates are igg(\frac{1}{2}; -\frac{9}{4}\bigg).

To derive the expression for length LM, we start with the linear equation of line L: $LM = \frac{1}{2}x + 2 - (x^2 + 3x)$ This simplifies to: $LM = -x^2 -\frac{5}{2}x + 2$ Completing the square results in: $LM = -\left( x + \frac{7}{4} \right)^2 + \frac{81}{16}$ Thus, the required form is achieved.