The graph of power functions

If the quadratic function is represented in vertex form,

f(x)=A(x-v)^2+B

we have seen that the parameters $A$ , $v$ and $B$ show how the reference function $r(x)=x^2$ has to be transformed to get to the graph of $f$ . The same applies to general power functions of the form

f(x)=A(x-v)^n+B

So the transformations we have to apply to get from the reference function $r(x)=x^n$ to the graph of $f$ ,

Stretch the graph of the reference function $r(x)=x^n$ by $A$ in the $y$ direction:
$x^2 \rightarrow A x^n$
Then move the graph to the left ( $v<0$ ) or to the right ( $v>0$ ) by $v$ :
$A x^n \rightarrow A(x-v)^n$
Then move the graph up ( $B>0$ ) or down ( $B<0$ ) by $B$ :
$A(x-v)^n \rightarrow A(x-v)^n+B$

The picture below shows an overview of the transformed functions $f(x)=A(x-v)^n+B$ for the different exponents $n$ . Note that we also indicate the point $S(v\vert B)$ , which is simply where the coordinate orgigin $(0|0)$ ends up when transformed. For $n=\text{2}$ , the quadratic function, this is also the vertex $S$ .

Example 1

Sketch the graph of the function

f(x)=2(x-1)^3-1

by transforming the reference graph. Also determine the $x$ - and $y$ intercepts.

Solution

It is $A=2, v=1, B=-1$ , i.e. stretching the reference function (black curve) $r(x)=x^3$ by a factor of $2$ in the $y$ direction, then shifting it $1$ to the right and $1$ downwards.

$y$ -intercept: $y=f(0)=2(-1)^3-1=\underline{-3}$ $x$ -intercept: find $x$ with

\begin{array}{rlll} 2(x-1)^3-1 &=& 0 &\quad\vert +1, :2\\ (x-1)^3 &= &0.5 &\quad\vert (.)^{1/3}\\ x-1 &=& 0.5^{1/3} &\quad\vert +1\\ x &=& 0.5^{1/3}+1 & \\ \end{array}

Thus, it is $x=\underline{1.793...}$ .

Example 2

Sketch the graph of the function

f(x)=\frac{3}{x+2}+1

by transforming the reference graph. Also determine the $x$ - and $y$ -intercepts.

Solution

Because of

\begin{array}{lll} f(x) &= &\frac{3}{x+2}+1\\ &=& 3\cdot \frac{1}{x+2}+1\\ &=& 3\cdot (x+2)^{-1}+1\\ \end{array}

follows $A=3, v=-2, B=1$ , thus stretching the reference function (green curve) $r(x)=\frac{1}{x}$ by the factor $3$ in $y$ -direction, then shift to the left by $2$ and upwards by $1$ .

$y$ -intercept: $y=f(0)=\frac{3}{x+2}+1=\underline{2.5}$ $x$ -intercept: find $x$ with

\begin{array}{rlll} \frac{3}{x+2}+1 &=& 0 &\quad\vert -1, \cdot(x+2)\\ 3 &= &-(x+2) &\\ 3 &=& -x-2 &\\ x &=& \underline{-5} & \\ \end{array}

Exercise 1

Q1

Sketch the graph of the power functions below by transforming the reference function. With the help of the sketch, determine whether the graph has $x$ - and $y$ -intercepts. If so, determine them mathematically.

$f(x)=\sqrt{x-1}+2$
$g(x)=-3(x+5)^4$
$h(x)=\frac{2}{x+1}+1$
$k(x)=3\cdot \sqrt[3]{8x-1}-4$

Q2

Determine the function equation of the two power functions below. The reference functions are $\sqrt{x}$ and $\frac{1}{x^2}$ .

Q3

Find the intersection between the graphs of the functions $f(x)=x^2$ and $g(x)=\frac{2}{\sqrt{x}}$ .

Solution

A1

Function $f$ : $f(x)=1\cdot (x-1)^{1/2}+2$ , thus $A=1, v=1, B=2$ . the reference function $r(x)=\sqrt{x}$ is not stretched in $y$ -direction, and shifted to the right by $1$ and upwards by $2$ . Clearly there is not -x- and -y-intercept.

Function $g$ : $g(x)=-3(x+5)^4+0$ , thus $A=-3, v=-5, B=0$ . The reference function $r(x)=x^4$ is stretched in $y$ -direction by $-3$ , that is, reflected about the $x$ -axis, and stretched by $3$ , and shifted to the left by $5$ . y-intercept: $g(0)=-3\cdot 5^4=\underline{-1875}$ x-intercept: $-3(x+5)^4=0$ , also $x+5=0$ , also $x=\underline{-5}$

Function $h$ : $h(x)=2\cdot (x+1)^{-1}+1$ , also $A=2, v=-1, B=1$ . The graph of the reference function $r(x)=\frac{1}{x}$ is stretched in $y$ -direction by the factor $2$ , shifted to the left by $1$ and upwards by $1$ . $y$ -intercept: $h(0)=\underline{3}$ $x$ -intercept:

\begin{array}{llll} 2(x+1)^{-1}+1 &=& 0 & \quad\vert -1, :2\\ (x+1)^{-1}&=& -\frac{1}{2} & \quad\vert (.)^{-1}\\ x+1 &=& \left(-\frac{1}{2}\right)^{-1} &\\ x &=& \underline{-3} & \end{array}

Function $k$ : Because of

\begin{array}{lll} k(x) &=& 3\cdot (8x-1)^{1/3}-4\\ &=& 3\cdot (8(x-\frac{1}{8}))^{1/3}-4\\ &=& 3\cdot 8^{1/3} \cdot (x-\frac{1}{8})^{1/3}-4\\ &=& 6 \cdot (x-\frac{1}{8})^{1/3}-4\\ \end{array}

follows $A=6, v=\frac{1}{8}, B=-4$ . The graph of the reference function $r(x)= \sqrt[3]{x}$ is stretched in $y$ -direction by the factor $6$ shifted to the right by $1/8$ and upwards by $4$ . $y$ -intercept: $k(0)=3\cdot (-1)^{1/3}-4 =-3-4=\underline{-7}$ . $x$ -intercept:

\begin{array}{llll} 3(8x-1)^{1/3}-4 &=& 0 & \quad\vert +4, :3\\ (8x-1)^{1/3}&=& \frac{4}{3} & \quad\vert (.)^{3}, +1\\ 8x &=& \left(\frac{4}{3}\right)^{3}+1 &\quad\vert :8\\ x &=& \underline{0.421} &\\ \end{array}

A2

Graph $f$ : The reference function is $r(x)=\sqrt{x}$ and from the picture we see that $r$ was shifted to the right by $2$ and downwards by $2$ . Thus, we have

f(x)=A\cdot (x-2)^{1/2}-2

To find $A$ , we use the fact that $f$ passes through the point $B(5\vert 1.9)$ , which means that

\begin{array}{llll} f(5) &=& 1.9 & \\ A\cdot (5-2)^{1/2}-2 &=& 1.9 &\quad\vert +2\\ \sqrt{3} A & = & 3.9 & \quad\vert :\sqrt{3}\\ A &=& 2.25... & \\ \end{array}

Thus we have $f(x)=\underline{2.25\cdot \sqrt{x-2}-2}$ .

Graph $g$ : The reference function is $r(x)=\frac{1}{x^2}$ and from the picture we see that $r$ was shifted to the left by $1$ and downwards by $4$ . Thus

g(x)=A(x+1)^{-2}-4

To find $A$ , we use the fact that $g$ passes through the point $B(0\vert-3.5)$ , thus

\begin{array}{llll} g(0) &=& -3.5 & \\ A\cdot (1)^{-2}-4 &=& -3.5 &\\ A & = & 0.5 & \\ \end{array}

It follows $g(x)=\underline{\frac{0.5}{(x+1)^2}-4}$ .

To find $A=0.5$ we could also compare the graphs of $r$ and $g$ .

A3

Find $x$ with $f(x)=g(x)$ :

\begin{array}{llll} x^2 &=& \frac{2}{\sqrt{x}}\\ x^2 \cdot \sqrt{x} &=& 2\\ x^{2.5} &=& 2 & \quad\vert (.)^{1/2.5}=(.)^{2/5}\\ x &=& 2^{2/5}\\ \end{array}

It is $x=\underline{1.32}$ . The $y$ -coordinate is therefore

y=f(1.32)=1.74

and the point of intersection is $\underline{P(1.32\vert 1.74)}$ .