Functions (Graphs and Behaviour)

Hong Kong

Stage 4 - Stage 5

Lesson

We first recall some basic graph types schematically shown in the diagram below.

If two linear functions of the form $y=mx+b$`y`=`m``x`+`b` (depicted on the far left) are drawn together on the one axes, they will intersect each other whenever their gradients differ.

As an example, the functions $y=12-3x$`y`=12−3`x` and $y=2x+2$`y`=2`x`+2 intersect at $\left(2,6\right)$(2,6). We know this is true because the point $\left(2,6\right)$(2,6) satisfies both equations. By 'satisfy' we mean that the equations become simultaneously true when we substitute $x=2$`x`=2 and $y=6$`y`=6 into each equation:

$6=12-3\left(2\right)$6=12−3(2)

$6=2\left(2\right)+2$6=2(2)+2

In particular, we need to recognises that the $y$`y` values are equal at the point of intersection.

Thus if $y=12-3x$`y`=12−3`x` and $y=2x+2$`y`=2`x`+2 are *equal* at this point, then it follows that $12-3x=2x+2$12−3`x`=2`x`+2. It is then a straight-forward process to solve for $x$`x`, revealing with a little algebra that $x=2$`x`=2. If $x=2$`x`=2, then *for both equations*, $y=6$`y`=6.

If a line given by $y=mx+b$`y`=`m``x`+`b` and a parabola given by the quadratic function $y=ax^2+bx+c$`y`=`a``x`2+`b``x`+`c` intersect, then they can do so in either one point (the line is a tangent to the parabola) or two points (the line is a secant to it).

Of course a line and a parabola need not intersect, so there are three possible scenarios to deal with.

The key understanding is that when we equate a line with equation $y=mx+b$`y`=`m``x`+`b` with a parabola with equation $y=px^2+qx+r$`y`=`p``x`2+`q``x`+`r` then we are solving for $x$`x` in the equation $px^2+qx+r=mx+b$`p``x`2+`q``x`+`r`=`m``x`+`b`. Gathering the like terms together onto one side of this last equation shows $px^2+\left(q-m\right)x+\left(r-b\right)=0$`p``x`2+(`q`−`m`)`x`+(`r`−`b`)=0.

In other words solving for the intersection of a line and a parabola is equivalent to solving a quadratic equation. Such an equation can have two distinct real roots, a single double real root, or no real roots at all. These three cases correspond to the line as a secant or a tangent or a line that misses the parabola entirely.

As an example consider the parabola given by $y=x^2-6x+11$`y`=`x`2−6`x`+11 and the lines given by $y=2x-1$`y`=2`x`−1, $y=2x-5$`y`=2`x`−5 and $y=2x-9$`y`=2`x`−9.

The line $y=2x-1$`y`=2`x`−1 will intersect $y=x^2-6x+11$`y`=`x`2−6`x`+11 when $x^2-6x+11=2x-1$`x`2−6`x`+11=2`x`−1. This means when $x^2-8x+12=0.$`x`2−8`x`+12=0. After factoring we have $\left(x-6\right)\left(x-2\right)=0$(`x`−6)(`x`−2)=0 and the line intersects as a secant at $x=2$`x`=2 and $x=6$`x`=6.

The line $y=2x-5$`y`=2`x`−5 will intersect $y=x^2-6x+11$`y`=`x`2−6`x`+11 when $x^2-6x+11=2x-5$`x`2−6`x`+11=2`x`−5. This means when $x^2-8x+16=0$`x`2−8`x`+16=0. After factoring we have $\left(x-4\right)^2=0$(`x`−4)2=0 and the line intersects as a tangent at $x=4$`x`=4.

The line $y=2x-9$`y`=2`x`−9 will intersect $y=x^2-6x+11$`y`=`x`2−6`x`+11 when $x^2-6x+11=2x-9$`x`2−6`x`+11=2`x`−9. This means when $x^2-8x+20=0$`x`2−8`x`+20=0. By completing the square we can show that this becomes $\left(x-4\right)^2+4=0$(`x`−4)2+4=0 which is impossible since the left hand side is always positive. Therefore the line misses the parabola.

The three situations are shown here:

Certain other curves can also intersect but you may require the use of technology to find precisely where they do so. We can estimate these intersections with carful plots of the curve.

Examples of these include exponential graphs with lines, parabolas and even higher powered polynomial functions.

Consider the following linear equations:

$y=5x-7$`y`=5`x`−7 and $y=-x+5$`y`=−`x`+5

Plot the lines of the two equations on the same graph.

Loading Graph...State the values of $x$

`x`and $y$`y`which satisfy both equations.$x$

`x`= $\editable{}$$y$

`y`= $\editable{}$

Consider the functions $f\left(x\right)=5x+5$`f`(`x`)=5`x`+5 and $g\left(x\right)=2^x+5$`g`(`x`)=2`x`+5.

Complete the table of values. Give your answers to one decimal if necessary.

$x$ `x`$4$4 $4.3$4.3 $4.5$4.5 $4.8$4.8 $5$5 $f\left(x\right)$ `f`(`x`)$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $g\left(x\right)$ `g`(`x`)$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Which value of $x$

`x`in the table is closest to the solution of $f\left(x\right)=g\left(x\right)$`f`(`x`)=`g`(`x`)?$4.3$4.3

A$5$5

B$4.5$4.5

C$4$4

D$4.8$4.8

EUsing the value of $x$

`x`found in the previous part, approximate the coordinates where the graphs of $f\left(x\right)$`f`(`x`) and $g\left(x\right)$`g`(`x`) intersect.