The diagram shows the graph of $y = \text{f}'(x)$, which passes through the origin and has turning points at $(2, 5)$ and $(6, 0)$. The graph of $y = \text{f}'(x)$ has a vertical asymptote $x = -1$ and a horizontal asymptote $y = 5$.

1. Identify the $x$-intercepts of the graph of $y = \text{f}''(x)$.

$y = \text{f}''(x)$ represents the gradient function of $y = \text{f}'(x)$.
At the turning points of $y = \text{f}'(x)$, the gradient is zero.
Given the turning points are at $(2, 5)$ and $(6, 0)$, we have $\text{f}''(2) = 0$ and $\text{f}''(6) = 0$.
Thus, the $x$-intercepts are $(2, 0)$ and $(6, 0)$.

2. Determine the equations of the asymptotes for $y = \text{f}''(x)$.

As $x \to \pm\infty$, $y = \text{f}'(x)$ approaches the horizontal asymptote $y = 5$. This means the curve flattens out, so its gradient approaches $0$. Thus, $y = 0$ is a horizontal asymptote for $y = \text{f}''(x)$.
The gradient of $y = \text{f}'(x)$ approaches $+\infty$ near the vertical asymptote $x = -1$, making $x = -1$ a vertical asymptote for $y = \text{f}''(x)$ as well.
Equations of asymptotes: $x = -1$ and $y = 0$.

3. Determine the sign and shape of $y = \text{f}''(x)$ in each interval.

• For $x < -1$: $\text{f}'(x)$ is increasing towards $+\infty$, so $\text{f}''(x) > 0$. It approaches $y=0$ as $x \to -\infty$ and goes to $+\infty$ as $x \to -1^-$.
• For $-1 < x < 2$: $\text{f}'(x)$ is increasing, so $\text{f}''(x) > 0$. It comes from $+\infty$, crosses the positive $y$-axis, and meets the $x$-axis at $(2, 0)$.
• For $2 < x < 6$: $\text{f}'(x)$ is decreasing, so $\text{f}''(x) < 0$. It reaches a local minimum and meets the $x$-axis at $(6, 0)$.
• For $x > 6$: $\text{f}'(x)$ is increasing, so $\text{f}''(x) > 0$. It reaches a local maximum and approaches $y=0$ as $x \to \infty$.

1. Identify the $x$-coordinates of the stationary points of $\text{f}(x)$

Stationary points occur where $\text{f}'(x) = 0$.
From the given graph of $y = \text{f}'(x)$, the graph intersects the $x$-axis at the origin and at the turning point $(6, 0)$.
Therefore, $\text{f}'(x) = 0$ at $x = 0$ and $x = 6$.

2. Determine the nature of the stationary point at $x = 0$

Just before $x = 0$, $\text{f}'(x) < 0$.
Just after $x = 0$, $\text{f}'(x) > 0$.
Since $\text{f}'(x)$ changes from negative to positive, there is a local minimum point at $x = 0$.

3. Determine the nature of the stationary point at $x = 6$

Just before $x = 6$, $\text{f}'(x) > 0$.
Just after $x = 6$, $\text{f}'(x) > 0$.
Since $\text{f}'(x)$ does not change sign (remains positive), there is a point of inflexion at $x = 6$.

1. Determine the gradient of the tangent at $x = 2$ from the given graph of $\mathrm{f}'(x)$.

From the given graph of $y = \mathrm{f}'(x)$, there is a turning point at $(2, 5)$. This means that at $x = 2$, the value of the derivative is $\mathrm{f}'(2) = 5$.
Therefore, the gradient of the tangent at $x = 2$ is $5$.

2. Find the coordinates of the point of tangency on the graph of $y = \mathrm{f}(x)$.

The tangent passes through the origin $(0, 0)$ and has a gradient of $m = 5$. Thus, the equation of the tangent line is $y = 5x$.
Since the tangent touches the graph of $y = \mathrm{f}(x)$ at $x = 2$, we can substitute $x = 2$ into the tangent equation to find the $y$-coordinate of the point of tangency:
$y = 5(2) = 10$
The point of contact on the curve is $(2, 10)$.

3. Determine the gradient of the normal and find its equation.

The gradient of the normal is the negative reciprocal of the gradient of the tangent:
$m_{\text{normal}} = -\frac{1}{\mathrm{f}'(2)} = -\frac{1}{5}$
Using the point $(2, 10)$, the equation of the normal is:
$y - 10 = -\frac{1}{5}(x - 2)$
$y = -\frac{1}{5}x + \frac{2}{5} + 10$
$y = -\frac{1}{5}x + \frac{52}{5}$