Describe fully a sequence of transformations which transforms the curve $y^2=x$ onto the curve $\dfrac{y^2}{4}=1-x$. A curve has equation $y=f(x)$, where $f(x)=\begin{cases}2\sqrt{1-x} &\text{for }0\le x<1,\\0 &\text{otherwise.}\end{cases}$ Sketch the curve of $y=f(x)$ for $-2\le x\le 2$, stating the coordinates of the end points and the coordinates of any points where the curve crosses the axes. Sketch the curve with equation $y=|f(x)-1|$ for $-2\le x\le 2$, stating the coordinates of the end points and the coordinates of any points where the curve crosses the axes.

1. Identify the scaling transformation required for the $y$ term.

Starting with $y^2=x$, we need to transform $y^2$ into $\dfrac{y^2}{4}$.
Replacing $y$ with $\dfrac{y}{2}$ gives:
$\left(\frac{y}{2}\right)^2 = x \implies \frac{y^2}{4} = x$
This corresponds to a scaling parallel to the $y$-axis with a scale factor of $2$.

2. Identify the sequence of reflection and translation required for the $x$ term.

We need to transform the right-hand side from $x$ to $1-x$.
First, replacing $x$ with $-x$ gives:
$\frac{y^2}{4} = -x$
This corresponds to a reflection in the $y$-axis.

Next, replacing $x$ with $(x - 1)$ gives:
$\frac{y^2}{4} = -(x - 1) \implies \frac{y^2}{4} = 1 - x$
This corresponds to a translation of $1$ unit in the positive $x$-direction (or translation by the vector $\binom{1}{0}$).

1. Find the coordinates of the end points of the domain.

The domain given is $-2 \le x \le 2$.
At $x = -2$, since $x$ is outside the range $0 \le x < 1$, the function falls under the 'otherwise' condition:
$f(-2) = 0$
At $x = 2$, similarly:
$f(2) = 0$
The end points are $(-2, 0)$ and $(2, 0)$.

2. Find the coordinates of any points where the curve crosses the axes.

For the $y$-axis intercept ($x = 0$):
Since $0$ is in the range $0 \le x < 1$, we use the first condition:
$f(0) = 2\sqrt{1 - 0} = 2$
The curve crosses the $y$-axis at $(0, 2)$.

For the $x$-axis, the graph lies flat along the axis for $-2 \le x < 0$ and $1 \le x \le 2$. There is no single discrete point where it 'crosses' the $x$-axis, other than resting on it.

3. Analyse the shape of the non-zero segment and describe the sketch.

For $0 \le x < 1$, the curve is $y = 2\sqrt{1-x}$. Squaring both sides gives $y^2 = 4(1-x)$, which is a parabola opening to the left. Since $y \ge 0$, it is the upper half of this parabola.
It starts at $(0, 2)$ and goes down to approach $f(1) = 0$.

To sketch:

1. Draw a horizontal line $y=0$ from $x=-2$ to $x=0$, ending with an open circle at $(0,0)$.
2. Draw a solid dot at $(0,2)$ and sketch the downward-concave curve from $(0,2)$ to $(1,0)$.
3. Draw a horizontal line $y=0$ from $x=1$ to $x=2$ (the curve continuously joins this line at $(1,0)$).

1. Determine the segments of the graph for intervals where the original function is zero.

For $-2 \le x < 0$ and $1 \le x \le 2$, we are given $f(x) = 0$.
The equation of the curve is $y = |f(x) - 1|$.
Substituting $f(x) = 0$, we get:
$y = |0 - 1| = 1$
Thus, the graph consists of horizontal line segments $y = 1$ for these intervals.

2. Determine the expression for the remaining interval and find the $y$-intercept.

For $0 \le x < 1$, $f(x) = 2\sqrt{1-x}$.
The curve is $y = \left|2\sqrt{1-x} - 1\right|$.
To find where it crosses the $y$-axis, substitute $x = 0$:
$y = \left|2\sqrt{1-0} - 1\right| = |2 - 1| = 1$
The curve crosses the $y$-axis at $(0, 1)$.

3. Find the $x$-intercept.

To find where the curve crosses the $x$-axis, set $y = 0$:
$\left|2\sqrt{1-x} - 1\right| = 0$
$2\sqrt{1-x} = 1$
$\sqrt{1-x} = \frac{1}{2}$
Squaring both sides:
$1 - x = \frac{1}{4} \implies x = \frac{3}{4}$
The curve crosses the $x$-axis at $(\frac{3}{4}, 0)$.

4. State the end points and verify the overall shape for the sketch.

To find the end points, evaluate $y$ at the boundaries $x = -2$ and $x = 2$:
At $x = -2$, $y = 1 \implies (-2, 1)$.
At $x = 2$, $y = 1 \implies (2, 1)$.

The sketch will feature:

• A horizontal line $y=1$ from $x=-2$ to $x=0$.
• A concave-down curve dropping from $(0,1)$ to touch the $x$-axis at a sharp point $V$ at $(\frac{3}{4},0)$.
• A concave-up curve rising from $(\frac{3}{4},0)$ to $(1,1)$.
• A horizontal line $y=1$ from $x=1$ to $x=2$.