How do you find the stretches of a transformed function?
Answer
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Hint: Assume a transformed function as $y=af\left( b\left( x-h \right) \right)+k$. Take two random values for ‘a’ and ‘b’. Find vertical stretch and horizontal stretch by using the formula $y'=ay$ and $x'=\dfrac{x}{b}$ respectively.
Complete step by step answer:
By multiplying a function by a coefficient, the graph of the function can be stretched.
Let a function given is $f\left( x \right)$ It can be stretched by following two ways:
The stretching of a function on x-axis is called vertical stretching .A function $g\left( x \right)$ represents a vertical stretch of $f\left( x \right)$ if $g\left( x \right)=cf\left( x \right)$ and $c>1$.
The stretching of a function on the y-axis is called horizontal stretching. A function $h\left( x \right)$ represents a horizontal stretch of $f\left( x \right)$ if $h\left( x \right)=f\left( cx \right)$ and $0 < c < 1$. (here ‘c’ is a variable of function $f\left( x \right)$ )
Let’s take an expression to find out the stretch factor of its variable.
$y=af\left( b\left( x-h \right) \right)+k$
In this expression, for $\left| a \right|>1$, the graph stretched vertically by a factor of ‘a’ unit and for $0 < \left| b \right| < 1$, the graph stretched horizontally by a factor of ‘b’ units.
Let’s take the value for both a and b.
Let a=4 and b=$\dfrac{1}{3}$
To find vertical stretch we can use the formula $y'=ay$
(where $y'$ represents the vertical stretch and ‘y’ represents the initial value)
So, $y'=4\times y$
Therefore vertical stretch would be by a factor of 4.
To find horizontal stretch we can use the formula $x'=\dfrac{x}{b}$
(where $x'$ represents the horizontal stretch and ‘x’ represents the initial value)
So,
$\begin{align}
& x'=\dfrac{x}{\dfrac{1}{3}} \\
& \Rightarrow x'~=3\times x \\
\end{align}$
Therefore vertical stretch would be by a factor of 3.
Note:
It should be marked that a vertical stretch coefficient is a number greater than 1 and a horizontal stretch coefficient is a number between 0 and 1. If $\left| a \right| < 1$, it will result in vertical compression. If $\left| b \right|>1$, it will result in horizontal compression.
Complete step by step answer:
By multiplying a function by a coefficient, the graph of the function can be stretched.
Let a function given is $f\left( x \right)$ It can be stretched by following two ways:
The stretching of a function on x-axis is called vertical stretching .A function $g\left( x \right)$ represents a vertical stretch of $f\left( x \right)$ if $g\left( x \right)=cf\left( x \right)$ and $c>1$.
The stretching of a function on the y-axis is called horizontal stretching. A function $h\left( x \right)$ represents a horizontal stretch of $f\left( x \right)$ if $h\left( x \right)=f\left( cx \right)$ and $0 < c < 1$. (here ‘c’ is a variable of function $f\left( x \right)$ )
Let’s take an expression to find out the stretch factor of its variable.
$y=af\left( b\left( x-h \right) \right)+k$
In this expression, for $\left| a \right|>1$, the graph stretched vertically by a factor of ‘a’ unit and for $0 < \left| b \right| < 1$, the graph stretched horizontally by a factor of ‘b’ units.
Let’s take the value for both a and b.
Let a=4 and b=$\dfrac{1}{3}$
To find vertical stretch we can use the formula $y'=ay$
(where $y'$ represents the vertical stretch and ‘y’ represents the initial value)
So, $y'=4\times y$
Therefore vertical stretch would be by a factor of 4.
To find horizontal stretch we can use the formula $x'=\dfrac{x}{b}$
(where $x'$ represents the horizontal stretch and ‘x’ represents the initial value)
So,
$\begin{align}
& x'=\dfrac{x}{\dfrac{1}{3}} \\
& \Rightarrow x'~=3\times x \\
\end{align}$
Therefore vertical stretch would be by a factor of 3.
Note:
It should be marked that a vertical stretch coefficient is a number greater than 1 and a horizontal stretch coefficient is a number between 0 and 1. If $\left| a \right| < 1$, it will result in vertical compression. If $\left| b \right|>1$, it will result in horizontal compression.
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