Transformation of Functions: Dilation (Stretches)

Vertical Stretches and Compressions

Given a function $f(x)$, a new function $g(x)=c f(x)$, where $c$ is a positive constant,
is a vertical stretch or vertical compression (parallel to the $y$-axis) of the function
$f(x)$ with a scale factor $c$.

• If $c>1$, then the graph will be stretched.




• If $0< c< 1$, then the graph will be compressed.

Horizontal Stretches and Compressions

Given a function $f(x)$, a new function $g(x)= f(cx)$, where $c$ is a positive constant,
is a horizontal stretch or horizontal compression (parallel to the $x$-axis) of the function
$f(x)$ with a scale factor $\frac{1}{c}$.

• If $c>1$, then the graph will be compressed by $\dfrac{1}{c}$.




• If $0< c< 1$, then the graph will be stretched $\dfrac{1}{c}$.

Question 1

The point $P(3, -2)$ lies on the graph $y=f(x)$.

State the coordinates of the map of the point $P$ on each of the following graph .


$\begin{array}{l}
\\
\text{(a) } y=f(2x)\\\\
\text{(b) } y=f(2x-1) \\\\
\text{(c) } y= f(2x-1) -3\\\\
\text{(d) } y=3f(x)\\\\
\text{(e) } y=-3f(x)\\\\
\text{(f) } y=-3f(-x)\\\\
\end{array}$
SOLUTION
Let the mapped point of $P(3, -2)$ be $P'(a, b)$.


$\begin{aligned}
&\\
\text{(a) } \quad & \text{ On the curve } y=f(2x),\\\\
&a=\dfrac{1}{2}\times 3 = \dfrac{3}{2}\\\\
&b=-2\\\\
\end{aligned}$


The mapped point of $P(3, -2)$ on the curve $y=f(2x)$ is $\left(\dfrac{3}{2},-2\right)$


$\begin{aligned}
&\\
\text{(b) } \quad & \text{ On the curve } y=f(2x-1),\\\\
&a=\dfrac{1}{2}\times (3 +1) = 2\\\\
&b=-2\\\\
\end{aligned}$


The mapped point of $P(3, -2)$ on the curve $y=f(2x)$ is $(2,-2)$


$\begin{aligned}
&\\
\text{(c) } \quad & \text{ On the curve } y=f(2x-1)-3,\\\\
&a=\dfrac{1}{2}\times (3 +1) = 2\\\\
&b=-2-3=-5\\\\
\end{aligned}$


The mapped point of $P(2, -5)$ on the curve $y=f(2x)$ is $(2,-5)$


$\begin{aligned}
&\\
\text{(d) } \quad & \text{ On the curve } y=3f(x),\\\\
&a=3\\\\
&b=3(-2)=-6\\\\
\end{aligned}$


The mapped point of $P(3, -2)$ on the curve $y=f(2x)$ is $(3,-6)$



$\begin{aligned}
&\\
\text{(e) } \quad & \text{ On the curve } y=-3f(x),\\\\
&a=3\\\\
&b=-3(-2)=6\\\\
\end{aligned}$


The mapped point of $P(3, -2)$ on the curve $y=f(2x)$ is $(3,6)$




$\begin{aligned}
&\\
\text{(d) } \quad & \text{ On the curve } y=-3f(-x),\\\\
&a=-3\\\\
&b=3(-2)=-6\\\\
\end{aligned}$


The mapped point of $P(3, -2)$ on the curve $y=f(2x)$ is $(-3,-6)$

Question 2

The point $P(-1, 3)$ lies on the graph $y=f(x)$. State the coordinates of the map of the point $P$
on each of the following graph .

$ \text{(a)}$	On the graph of $y=a f(x)$, where $a$ is a positive constant, the point $P$ is mapped to the point $(-1,1)$. Determine the value of $a$.
$ \text{(b)}$	On the graph of $y={f}(k x)$, where $k$ is a positive constant, the point $P$ is mapped to the point $(-3,3)$. Determine the value of $k$.

SOLUTION

$ \text{(a)}$	The point $P(−1,3)$ lies on the graph $y=f(x)$.
	The mapped point of $P$ on the graph $y=af(x)$ is $(-1, 3a)$.
	By the problem, the mapped point of $P$ on the graph $y=af(x)$ is $(-1, 1)$.
	$\therefore\ 3a = 1\Rightarrow a=\dfrac{1}{3}$
$ \text{(b)}$	The point $P(−1,3)$ lies on the graph $y=f(x)$.
	The mapped point of $P$ on the graph $y=f(kx)$ is $\left(-\dfrac{1}{k}, 3\right)$.
	By the problem, the mapped point of $P$ on the graph $y=af(x)$ is $(-3, 3)$.
	$\therefore\ -\dfrac{1}{k} = -3\Rightarrow k=\dfrac{1}{3}$

Question 3

The point $A(a, b)$ lies on the graph $y=f(x)$. State the coordinates of the map
of the point $A$ on each of the following graphs.


$\begin{array}{l}
\\
\text{(a) } y=\dfrac{1}{2}f(x) \\\\
\text{(b) } y=3f(x) + 1 \\\\
\text{(c) } y=f(2x-1)
\end{array}$

SOLUTION
The point $A(a,b)$ lies on the graph $y=f(x)$.


$\begin{aligned}
&\\
\text{(a) }\ &\text{On the curve } y=\dfrac{1}{2}f(x)\\\\
&\text{The mapped point of } A \text{ is } \left(a,\dfrac{b}{2}\right)\\\\
\text{(b) }\ &\text{On the curve } y=3f(x)+1\\\\
&\text{The mapped point of } A \text{ is } \left(a,3b+1\right)\\\\
\text{(c) }\ &\text{On the curve } y=f(2x-1)\\\\
&\text{The mapped point of } A \text{ is } \left(\dfrac{a+1}{2},b\right)\\\\
\end{aligned}$

Question 4

The figure shows the parabola with equation y = f ( x ) with vertex at B( 2,4).

The graph cuts the x axis at O( 0, 0) and at the point A( 4, 0).







Sketch on a separate set of axes the graph of …


$\begin{aligned}
&\\
&\text{(a) } y=3f(x)\\\\
&\text{(b) } y=f\left(\frac{x}{2}\right)\\\\
\end{aligned}$


Each sketch must include the coordinates of any points where the graph crosses the
coordinate axes and the new coordinates of the vertex of the curve.

SOLUTION



(a)


(b)

Question 5

The figure shows a sketch of the curve with equation $y = f(x)$.



The curve passes through the points $(0, 3)$ and $(4, 0)$ and touches
the $x$-axis at the point $(1, 0)$.







On separate diagrams sketch the curve with equation ...


$\begin{aligned}
&\\
&\text{(a) } y = f(2x)\\\\
&\text{(b) } y = 2f(x)\\\\
&\text{(c) } y = -2f(x)\\\\
\end{aligned}$


On each diagram show clearly the coordinates of all the points where the curve meets
the axes.

SOLUTION
(a)





(b)





(c)

Question 6

Given that $f(x)=\sqrt{x}$.



The following table represents the coordinate of some points on the graph $y=f(x)$.



$\begin{array}{|c|c|c|c|c|c|}
\hline x & 0 & 1 & 4 & 9 \\
\hline y & 0 & 1 & 2 & 3 \\
\hline
\end{array}$

$ \text{(a)}$	State the domain and range of $f(x)$.
$ \text{(b)}$	Use table to sketch the graph of $y=f(x)$.
$ \text{(c)}$	Describe the transformation that transforms the graph $y=\sqrt{x}$ to the graph $y=\sqrt{x-4}$.
$ \text{(d)}$	The graph $y=\sqrt{x}$ is stretched by a scale factor of $2$ parallel to the $x$-axis and then it is translated $3$ units right. State the equation of the transformed graph.
$ \text{(e)}$	The graph $y=\sqrt{x}$ is stretched by a scale factor of $5$ parallel to the $x$-axis. State the equation of the transformed graph.
$ \text{(f)}$	If $(8,4)$ is a point on the graph $y=f(cx)$, what is the value of $c$?

SOLUTION




$\textbf{(a)}$ domain of $f=\{x\mid x\ge 0, x\in\mathbb{R}\}$


$\quad\ \ $ range of $f= \{y\mid y\ge 0, y\in\mathbb{R}\}$.




$\textbf{(b)}$





$\textbf{(c)}$ The graph of $y=\sqrt{x-4}$ is obtained from $y=\sqrt{x}$ by shifting $4$ units right.




$\textbf{(d)}$ The equation of the transformed graph is $y=\sqrt{\dfrac{1}{2}(x-3)}$



$\textbf{(e)}$ The equation of the transformed graph is $y=\sqrt{\dfrac{1}{5}x}$




$\textbf{(f)}$ The coordinates of any point on the graph $y=f(cx)$ is $(x, \sqrt{cx})$.



By the problem, $(8, 4)$ lies on the graph $y=f(cx)$



$\begin{aligned}
\therefore\ \sqrt{8c} &= 4\\\\
8c &= 16\\\\
c&=2
\end{aligned}$

Berbagi