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 1 - 2016

To verify this, we evaluate $f(-5)$:

$f(-5) = (-5)^3 + 3(-5)^2 - 9(-5) + 5$
First, calculate:

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

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

To find the turning points, we first differentiate $f(x)$:

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

Setting the derivative equal to zero:

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

We simplify this to:

$x^2 + 2x - 3 = 0$
Factoring gives:

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

Thus, $x = 1$ and $x = -3$. To determine the nature of these points, we substitute back into the function:

For $x = -3$: $f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 5 = 32$

For $x = 1$: $f(1) = (1)^3 + 3(1)^2 - 9(1) + 5 = -6$

The local maximum turning point is $(-3, 32)$ and the local minimum turning point is $(1, -6)$.

Based on our calculations, we sketch the graph with local maxima and minima:

• The y-intercept is at (0, 5).
• It cuts the x-axis at (-5, 0).
• A local maximum occurs at (-3, 32) and a local minimum at (1, -6).

The general shape will reflect these critical points.