The function $f$ is defined as $f: \mathbb{R} \rightarrow x^3 + 3x^2 - 9x + 5$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question Question 1 - 2014

To verify if the graph cuts the $x$-axis, we need to evaluate $f$ at $x = -5$:

$f(-5) = (-5)^3 + 3(-5)^2 - 9(-5) + 5$

Calculating:

$f(-5) = -125 + 75 + 45 + 5 = 0$

Since $f(-5) = 0$, the graph does cut the $x$-axis at $x = -5$.

To find the turning points, we need to calculate the first derivative:

$f'(x) = 3x^2 + 6x - 9$

Setting the first derivative to zero gives:

$3x^2 + 6x - 9 = 0 \\ x^2 + 2x - 3 = 0$

Factoring, we find:

$(x + 3)(x - 1) = 0$

Thus, $x = -3$ and $x = 1$.

Next, we check the second derivative:

$f''(x) = 6x + 6$

For $x = -3$:

$f''(-3) = 6(-3) + 6 = -12 < 0 \\ \Rightarrow \text{ Local Maximum at } (-3, f(-3))$

Calculating $f(-3)$:

$f(-3) = 32 \Rightarrow \text{ Local Maximum is } (-3, 32)$

For $x = 1$:

$f''(1) = 6(1) + 6 = 12 > 0 \\ \Rightarrow \text{ Local Minimum at } (1, f(1))$

Calculating $f(1)$:

$f(1) = 1$

Thus, the Local Minimum is $(1, 1)$.

Based on the calculations:

• The $y$-intercept is at $(0, 5)$.
• The graph crosses the $x$-axis at $(-5, 0)$.
• The local maximum turning point is $(-3, 32)$ and the local minimum is $(1, 1)$.

Use this information to sketch the graph accurately on the provided axes.