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Stage 4 - Stage 5

# Intersections of graphs by other methods II

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=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 $\left(m-p\right)x+\left(b-q\right)=0$(mp)x+(bq)=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=bqmp=qbmp

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=x2.

This can be found by setting $2x+3=x-2$2x+3=x2. Gathering like terms together onto one side shows $\left(2-1\right)x+\left(3--2\right)=0$(21)x+(32)=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(2xx) 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(2xx) 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 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$1x1=x22+1.

1. One of the graphs is $f\left(x\right)=\frac{1}{x-1}$f(x)=1x1. What function $g\left(x\right)$g(x) should the other graph be of?

2. The graph of $f\left(x\right)$f(x) has been provided for $x\ge0$x0. On the same set of axes, graph $g\left(x\right)=\frac{x^2}{2}+1$g(x)=x22+1.

3. 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.

1. 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).

2. 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

D
3. You 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=x0x30+2x20x043x20+4x01 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.