The figure below shows the sketch of the graph $y = f(x)$. The graph has asymptotes at $x = 1$, $y = 3$ and the x-axis. The graph has stationary point at $(-1,-3)$ and cuts the axes at $(0,0)$ and $(2,0). Sketch, on separate diagrams, the following graphs, indicating clearly any asymptotes, axial intercepts and stationary points, where possible. Sketch $y = -2f(x) + 1$ Sketch $y = \dfrac{1}{f(x)}$ Sketch $y = f(|x| + 1)$ Describe fully a sequence of transformations that transform the graph of $y = x^2$ onto the graph of $y = 4x^2 - 4x + 1$.

1. Apply the vertical transformations to find the new coordinates of the known key points.

The transformation $y = -2f(x) + 1$ affects only the $y$-coordinates of the graph. It scales the graph vertically by a factor of 2, reflects it across the $x$-axis, and translates it upwards by 1 unit. We apply $y_{new} = -2y_{old} + 1$ to the given points:

• Stationary point $(-1, -3) \rightarrow (-1, -2(-3) + 1) = (-1, 7)$. Since it was a minimum (trough), the reflection makes it a maximum.
• Origin / $x$-intercept $(0, 0) \rightarrow (0, -2(0) + 1) = (0, 1)$. This gives the new $y$-intercept.
• $x$-intercept $(2, 0) \rightarrow (2, -2(0) + 1) = (2, 1)$.

2. Determine the new equations of the asymptotes.

Apply the same transformation rules to the horizontal asymptotes. The vertical asymptote is unaffected by vertical transformations.

• Vertical asymptote: remains $x = 1$.
• Horizontal asymptote $y = 0$ (for $x < 0$): $y_{new} = -2(0) + 1 = 1$.
• Horizontal asymptote $y = 3$ (for $x > 1$): $y_{new} = -2(3) + 1 = -5$.

3. Determine the shape of the branches and sketch the graph.

Analyse the behaviour near asymptotes to ensure the correct shape:

• Left branch ($x < 1$): The original graph approached $y=0$ from below as $x \to -\infty$. The new graph will approach $y=1$ from above. It rises to the maximum at $(-1, 7)$, falls through $(0, 1)$, and since the original rose steeply as $x \to 1^-$, the new graph drops steeply to $-\infty$ as $x \to 1^-$.
• Right branch ($x > 1$): The original dropped to $-\infty$ as $x \to 1^+$. The new graph will rise from $+\infty$. It falls through $(2, 1)$ and since the original approached $y=3$ from below as $x \to \infty$, the new curve approaches $y=-5$ from above.
• Combine these properties to draw the full sketch.

1. Find the vertical asymptotes and axis intercepts of the new graph.

The $x$-intercepts of $y = f(x)$ become the vertical asymptotes of $y = \frac{1}{f(x)}$.
$f(0) = 0 \text{ and } f(2) = 0 \implies \text{Vertical asymptotes at } x = 0 \text{ and } x = 2$

The vertical asymptote of $y = f(x)$ becomes an $x$-intercept for $y = \frac{1}{f(x)}$.
$f(x) \to \pm\infty \text{ as } x \to 1 \implies x\text{-intercept at } (1, 0)$

2. Find the horizontal asymptote and the coordinates of the stationary point.

The horizontal asymptote is found by taking the reciprocal of the limit as $x \to \infty$:
$f(x) \to 3 \text{ as } x \to \infty \implies y \to \frac{1}{3}$
So, there is a horizontal asymptote at $y = \frac{1}{3}$.

The local minimum of $f(x)$ becomes a local maximum on the reciprocal graph. Taking the reciprocals of the $y$-coordinates:
$\text{Local minimum at } (-1, -3) \implies \text{Local maximum at } \left(-1, -\frac{1}{3}\right)$

3. Determine the shape of the curve in each region to complete the sketch.

Analyse the behaviour of $\frac{1}{f(x)}$ based on the sign and magnitude of $f(x)$:

• For $x < 0$: The x-axis is an asymptote for $f(x)$, so $f(x) \to 0^-$ as $x \to -\infty$. Thus, $y \to -\infty$. The curve rises to the maximum at $\left(-1, -\frac{1}{3}\right)$ and goes to $-\infty$ as $x \to 0^-$.
• For $0 < x < 2$: The curve continuously drops from $+\infty$ (at $x=0^+$) to $-\infty$ (at $x=2^-$), passing through the $x$-axis at $(1,0)$.
• For $x > 2$: Since $f(x)$ approaches 3 from below and passes through $(2,0)$, it is positive. Hence $y \to +\infty$ as $x \to 2^+$ and decreases to approach $y = \frac{1}{3}$ from above as $x \to \infty$.

1. Identify the relevant domain of the original graph and the symmetry of the new function.

For $y = \mathrm{f}(|x| + 1)$, since $|x| \ge 0$ for all real $x$, the argument evaluates to $|x| + 1 \ge 1$.
This means the new graph is generated purely from the portion of $y = \mathrm{f}(x)$ where $x \ge 1$. The parts of $\mathrm{f}(x)$ for $x < 1$ (including the intercept at the origin and the stationary point at $(-1, -3)$) are excluded entirely.

Additionally, replacing $x$ with $-x$ gives $\mathrm{f}(|-x| + 1) = \mathrm{f}(|x| + 1)$. Since the function is even, its graph is perfectly symmetrical about the $y$-axis.

2. Determine the new axial intercepts.

To find the $x$-intercepts, set $y = 0 \implies \mathrm{f}(|x| + 1) = 0$.
From the given graph, the original $x$-intercepts are at $x = 0$ and $x = 2$.
$|x| + 1 = 0 \implies |x| = -1$ (no real solution).
$|x| + 1 = 2 \implies |x| = 1 \implies x = 1$ or $x = -1$.
Thus, the new $x$-intercepts are $(-1, 0)$ and $(1, 0)$.
There is no $y$-intercept because $x = 0$ corresponds to evaluating $\mathrm{f}(1)$, which lies on a vertical asymptote.

3. Determine the new asymptotes.

Vertical asymptote:
The original graph has a vertical asymptote at $x = 1$.
$|x| + 1 = 1 \implies |x| = 0 \implies x = 0$.
As $x \to 0$, $|x| + 1 \to 1^+$. The description states that the original curve drops to large negative values just to the right of $x = 1$. Therefore, the new graph drops to $-\infty$ on both sides of the $y$-axis. The $y$-axis ($x = 0$) is the vertical asymptote.

Horizontal asymptote:
As $x \to \pm \infty$, $|x| + 1 \to +\infty$.
The original graph approaches $y = 3$ as $x \to +\infty$.
Thus, $y = 3$ remains the horizontal asymptote on both sides.

1. Rewrite the target equation in a form that clearly shows the individual transformations from $y = x^2$.

$y = 4x^2 - 4x + 1$
$y = (2x - 1)^2$
$y = \left[2\left(x - \frac{1}{2}\right)\right]^2$
$y = 4\left(x - \frac{1}{2}\right)^2$

2. State the sequence of transformations by relating the modified equation back to the base curve.

Comparing $y = 4\left(x - \frac{1}{2}\right)^2$ to the base graph $y = x^2$, the required transformations are:

1. Replace $x$ with $\left(x - \frac{1}{2}\right)$: A translation of $\frac{1}{2}$ unit in the positive $x$-direction, which gives $y = \left(x - \frac{1}{2}\right)^2$.
2. Multiply the function by $4$: A scaling parallel to the $y$-axis by a scale factor of $4$, which gives $y = 4\left(x - \frac{1}{2}\right)^2$.