Functions (Graphs and Behaviour)

Hong Kong

Stage 4 - Stage 5

Lesson

We recall that the points of intersection of two curves $y=f\left(x\right)$`y`=`f`(`x`) and $y=g\left(x\right)$`y`=`g`(`x`) are found by setting $f\left(x\right)=g\left(x\right)$`f`(`x`)=`g`(`x`).

For example the single intersection point of the two lines $y=mx+b$`y`=`m``x`+`b` and $y=px+q$`y`=`p``x`+`q` can be found by setting $mx+b=px+q$`m``x`+`b`=`p``x`+`q`. Gathering like terms together onto one side shows $\left(m-p\right)x+\left(b-q\right)=0$(`m`−`p`)`x`+(`b`−`q`)=0 and thus the $x$`x`- value of point we are searching for is given by $x=-\frac{b-q}{m-p}=\frac{q-b}{m-p}$`x`=−`b`−`q``m`−`p`=`q`−`b``m`−`p`.

Note that this last result is *not* a formula to be learnt, but rather a concept to be understood. We learn two things from this algebra. Firstly, the lines do not intersect if $m=p$`m`=`p`, and secondly there is only one possible point of intersection between any two two lines.

But we can put it into perspective through and example.

Find the single intersection point of the two lines $y=2x+3$`y`=2`x`+3 and $y=x-2$`y`=`x`−2.

This can be found by setting $2x+3=x-2$2`x`+3=`x`−2. Gathering like terms together onto one side shows $\left(2-1\right)x+\left(3--2\right)=0$(2−1)`x`+(3−−2)=0 and thus the $x$`x`-value of point we are searching for is given by $x=-5$`x`=−5.

We can extend the same idea to higher degree polynomials. A quadratic function $y=px^2+qx+r$`y`=`p``x`2+`q``x`+`r` and a line $y=mx+b$`y`=`m``x`+`b` can only intersect in at most two places, because there can only ever be two solutions to the equation $ax^2+bx+c=mx+b$`a``x`2+`b``x`+`c`=`m``x`+`b`. The same reasoning informs us that there can never be any more than $3$3 intersections between the cubic function and the quadratic function.

With other function forms however, this is not so clear. For example, the exponential curve given by $y=\frac{1}{2}\left(2^x-x\right)$`y`=12(2`x`−`x`) and the absolute value function given by $y=\left|x+1\right|$`y`=|`x`+1| intersect in three points as shown in this graph:

The intersection points were found using computer technology, and are given as approximately $\left(-2.215,1.215\right)$(−2.215,1.215), $\left(-0.417,0.583\right)$(−0.417,0.583) and $\left(3.717,4.717\right)$(3.717,4.717).

There are various clever mathematical methods that the technology uses to find these intersections. One such method is called the *Newton-Raphson method* which involves concepts in the calculus.

We can however use a *plotting strategy* to locate the approximate positions of intersections.

Without actually graphing the functions, we could determine, say, $11$11 function values, for *both* functions, for integer values of $x$`x` from say $x=-5$`x`=−5 to $x=5$`x`=5.

This table shows the $5$5 function values for $x=1,2,3,4,5$`x`=1,2,3,4,5 on both functions:

x value | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

$y=\frac{1}{2}\left(2^x-x\right)$y=12(2x−x) |
0.5 | 1 | 2.5 | 6 | 13.5 |

$y=\left|x+1\right|$y=|x+1| |
2 | 3 | 4 | 5 | 6 |

Looking at the table only, the relative sizes of the $y$`y` - values of the two functions change around between $x=3$`x`=3 and $x=4$`x`=4. This tells us that there is an intersection point within the interval $3.$3.

To progress further we could divide this shorter interval up and check again for a change in relative sizes. Continued application of this process can lead to better and better approximations of the points of intersections.

Xavier is going to use the point of intersection of graphs to find an approximation to the positive solution of $\frac{1}{x-1}=\frac{x^2}{2}+1$1`x`−1=`x`22+1.

One of the graphs is $f\left(x\right)=\frac{1}{x-1}$

`f`(`x`)=1`x`−1. What function $g\left(x\right)$`g`(`x`) should the other graph be of?The graph of $f\left(x\right)$

`f`(`x`) has been provided for $x\ge0$`x`≥0. On the same set of axes, graph $g\left(x\right)=\frac{x^2}{2}+1$`g`(`x`)=`x`22+1.Loading Graph...Between which two values does the solution of the equation lie? There may be more than one solution.

$x=-2$

`x`=−2 and $x=3$`x`=3A$x=2$

`x`=2 and $x=3$`x`=3B$x=1$

`x`=1 and $x=2$`x`=2C$x=0$

`x`=0 and $x=1$`x`=1D

Consider the functions $f\left(x\right)=x^3+2x^2$`f`(`x`)=`x`3+2`x`2 and $g\left(x\right)=x+4$`g`(`x`)=`x`+4.

The graph of $f\left(x\right)$

`f`(`x`) has been drawn. On the same set of axes, graph the function $g\left(x\right)$`g`(`x`).Loading Graph...Which of the following would be the closest approximation to the solution of $x^3+2x^2=x+4$

`x`3+2`x`2=`x`+4?$x=3$

`x`=3A$x=1$

`x`=1B$x=-1$

`x`=−1C$x=0$

`x`=0DYou can use the formula $x_1=x_0-\frac{x_0^3+2x_0^2-x_0-4}{3x_0^2+4x_0-1}$

`x`1=`x`0−`x`30+2`x`20−`x`0−43`x`20+4`x`0−1 to find a closer approximation, where $x_0$`x`0 is an initial approximation. Using the result of the previous part as an initial approximation, determine the closer approximation $x_1$`x`1 to one decimal place.