Intersections of graphs by other methods II
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=mx+b and $y=px+q$y=px+q can be found by setting $mx+b=px+q$mx+b=px+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−qm−p=q−bm−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.
Example 1
Find the single intersection point of the two lines $y=2x+3$y=2x+3 and $y=x-2$y=x−2.
This can be found by setting $2x+3=x-2$2x+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=px2+qx+r and a line $y=mx+b$y=mx+b can only intersect in at most two places, because there can only ever be two solutions to the equation $ax^2+bx+c=mx+b$ax2+bx+c=mx+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(2x−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.
Worked Examples
QUESTION 1
QUESTION 2
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$1x−1=x22+1.
One of the graphs is $f\left(x\right)=\frac{1}{x-1}$f(x)=1x−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)=x22+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=3
A$x=2$x=2 and $x=3$x=3
B$x=1$x=1 and $x=2$x=2
C$x=0$x=0 and $x=1$x=1
D
QUESTION 3
Consider the functions $f\left(x\right)=x^3+2x^2$f(x)=x3+2x2 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 $x^3+2x^2=x+4$x3+2x2=x+4?
$x=3$x=3
A$x=1$x=1
B$x=-1$x=−1
C$x=0$x=0
DYou can use the formula $x_1=x_0-\frac{x_0^3+2x_0^2-x_0-4}{3x_0^2+4x_0-1}$x1=x0−x30+2x20−x0−43x20+4x0−1 to find a closer approximation, where $x_0$x0 is an initial approximation. Using the result of the previous part as an initial approximation, determine the closer approximation $x_1$x1 to one decimal place.