Transformations of Square Root Functions

 

Transformations on $y=\sqrt{x}$y=√x

The basic root function can be dilated and translated in a similar way to other functions.

The root function $y=\sqrt{x}$y=√x can be transformed to $y=a\sqrt{x-h}+k$y=a√x−h+k by dilating it using the factor $a$a and translating it, first horizontally $h$h units to the right, and then $k$k units upward. If the factor $a$a is negative, the basic curve reflects across the $x$x axis.

The subsequent translation then takes this reflected curve and moves the point at the origin to the point $\left(h,k\right)$(h,k).

A word on notation

Strictly speaking, we really should write that if the original function $y_1=\sqrt{x}$y1​=√x is dilated by a factor $a$a and then translated right by $h$h and up by $k$k, then the translated function becomes $y_2=a\sqrt{x-h}+k$y2​=a√x−h+k. Most times though we just label both functions $y$y, because we understand that the basic function is different to the transformed function. 

Using function notation, if $f\left(x\right)=\sqrt{x}$f(x)=√x then, for example, after a dilation of $2$2, and translations of  $5$5 units to the right and $3$3 units down, we create a new transformed function, say $g\left(x\right)$g(x) ,where:

$g\left(x\right)=2\times f\left(x-5\right)-3=2\sqrt{x-5}-3$g(x)=2×f(x−5)−3=2√x−5−3

Applet play

The best thing to do is to experiment with the first applet below showing how the variables $a,h$a,h and $k$k change the basic curve. Try both negative and positive values of $a$a.  

Focussing on the domain and range of the transformed function $y=a\sqrt{x-h}+k$y=a√x−h+k, note that we need $x-h\ge0$x−h≥0 and thus $x\ge h$x≥h.

The new range, because of the lift (or fall) caused by $k$k,  has also changed to the interval $k\le y<\infty$k≤y<∞ for values of $a>0$a>0. If however $a$a is negative the new range becomes $-\infty−∞<y≤k. Experiment with the applet, taking careful note of the natural domains and ranges.

The impact of transformations on nth root functions 

We can extend our idea of root functions to include $n$nth roots rather than just square roots.

The nth root of x can be written as $x^{\frac{1}{n}}$x1n​, where n is a positive integer greater than $1$1. The function $f\left(x\right)=\sqrt[n]{x}=x^{\frac{1}{n}}$f(x)=ⁿ√x=x1n​ is defined for positive $x$x irrespective of the parity (oddness or evenness) of $n$n . However it is defined for negative $x$x  only if $n$n is odd.

As an example, the domain for $f\left(x\right)=\sqrt[3]{x}$f(x)=³√x includes all real numbers, but the domain for $f\left(x\right)=\sqrt[4]{x}=x^{\frac{1}{4}}$f(x)=⁴√x=x14​ is only defined for $x\ge0$x≥0. The reason that values of $x$x cannot be negative when $n$n is $4$4 (or indeed any even number), is because numbers like $\sqrt[4]{-1}$⁴√−1 , $\sqrt[4]{-3.5}$⁴√−3.5 etc. are not real. 

This second applet, illustrating the general root function $y=a\sqrt[n]{x-h}+k$y=aⁿ√x−h+k, demonstrates this very well. It has the option of dilating and translating the basic root function just like the first applet, but also has a slider to increase $n$n. As you play with the sliders in combination, it is important that you record your new learnings somewhere.

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3