Functions (Graphs and Behaviour)

Hong Kong

Stage 4 - Stage 5

Lesson

Let's first recall the basic graphs as shown here in the diagram. The function's description is given below each graph. For example the linear function $y=mx+b$`y`=`m``x`+`b` is shown on the left and the hyperbola $y=\frac{1}{x}$`y`=1`x` is shown second from right.

Two of the graphs are shown with some generalising coefficients - the line $y=mx+b$`y`=`m``x`+`b` and the exponential curve $y=a^x$`y`=`a``x`. Thus we could site particular examples of these graphs such as $y=2x+5$`y`=2`x`+5, $y=7x$`y`=7`x` and $y=3^x$`y`=3`x` etc.

However, all of these basic types can be generalised further by what are known as *transformations*.

One common transformation is known as a *translation*. This is a horizontal and/or vertical shift in the position of the curve relative to the cartesian axes. Translations are easy to recognise.

If we replace $x$`x` with $\left(x-h\right)$(`x`−`h`), the basic curve is moved $h$`h` units to the right. If we replace $x$`x` with $\left(x+h\right)$(`x`+`h`), the basic curve is moved $h$`h` units to the left. If we replace $y$`y` with $\left(y-k\right)$(`y`−`k`), the basic curve is moved $k$`k` units up. If we replace $y$`y` with $\left(y+k\right)$(`y`+`k`), the basic curve is moved $k$`k` units down.

So suppose we take the hyperbola $y=\frac{1}{x}$`y`=1`x` and translate it $3$3 units to the right and $5$5 units down. The new function becomes $y+5=\frac{1}{x-3}$`y`+5=1`x`−3, or expressed explicitly $y=\frac{1}{x-3}-5$`y`=1`x`−3−5. The vertical and horizontal asymptotes are respectively $x=3$`x`=3 and $y=-5$`y`=−5.

If we take the parabola $y=x^2$`y`=`x`2 and shift it $1$1 unit to the left and $5$5 units up, the new function becomes $y-5=\left(x-1\right)^2$`y`−5=(`x`−1)2 . Again, we can express this explicitly as $y=\left(x-1\right)^2+5$`y`=(`x`−1)2+5.

If we take the function $y=x^3$`y`=`x`3 and translate it $2$2 units to the right and $1$1 unit up, the new function is given by $y-1=\left(x-2\right)^3$`y`−1=(`x`−2)3, expressible as $y=\left(x-2\right)^3+1$`y`=(`x`−2)3+1.

Another type of transformation is commonly referred to as a dilation. This is when a curve is stretched or compressed by some factor other than $1$1 in the function's equation.

For example, the difference between $y=x^2$`y`=`x`2 to $y=3x^2$`y`=3`x`2 is the dilation factor $3$3. Every function value in $y=3x^2$`y`=3`x`2 is $3$3 times the associated function value in $y=x^2$`y`=`x`2. This means that the curve becomes steeper.

Again, every function value of the function $y=\frac{2}{x}$`y`=2`x` is double the associated function value of $y=\frac{1}{x}$`y`=1`x`.

Every function value of $y=3\left(2^x\right)$`y`=3(2`x`) is $3$3 times that of $y=2^x$`y`=2`x`, and every function value of $y=\frac{1}{2}x^3$`y`=12`x`3 is one-halve that of $y=x^3$`y`=`x`3.

We can also reflect a curve across the $x$`x` - axis by simply multiplying its function expression by $-1$−1.

Every function value of $y=-x^2$`y`=−`x`2 is the negative of those of $y=x^2$`y`=`x`2, and the curve becomes an upside-down parabola with the same vertex as $y=x^2$`y`=`x`2. Again, every function value of $y=-\frac{3}{x}$`y`=−3`x` is the reflected image of across the x -axis of $y=\frac{3}{x}$`y`=3`x`.

When we combine translations and dilations we can reposition and distort the basic curve.

For example, the curve of $y=\frac{5}{x-2}+3$`y`=5`x`−2+3 is constructed as follows. Start with the curve given by the basic function $y=\frac{1}{x}$`y`=1`x` and stretch each ordinate by a factor of $5$5. Then shift it $2$2 units to the right and $3$3 units up, so that the centre becomes $\left(2,3\right)$(2,3) with the curve's asymptotes as $x=2$`x`=2 and $y=3$`y`=3.

The final graph is shown in red here, along with the graph of $y=\frac{1}{x-2}+3$`y`=1`x`−2+3. Note how changing from $y=\frac{1}{x-2}+3$`y`=1`x`−2+3 to $y=\frac{5}{x-2}+3$`y`=5`x`−2+3 pushes the hyperbola out diagonally.

The $y$`y` intercept of any function are found by putting $x=0$`x`=0 into the equation. The $x$`x` intercept(s) are found similarly by putting $x=0$`x`=0.

For example, for the curve given by $y=\frac{1}{x-3}-5$`y`=1`x`−3−5, putting $x=0$`x`=0 reveals $y=\frac{1}{0-3}-5=-5\frac{1}{3}$`y`=10−3−5=−513.

The x intercept is found by putting $y=0$`y`=0, so that $\frac{1}{x-3}-5=0$1`x`−3−5=0. This means that $\frac{1}{x-3}=5$1`x`−3=5 and inverting $x-3=\frac{1}{5}$`x`−3=15 so that $x=3\frac{1}{5}$`x`=315.

Consider the equation $y=-2x$`y`=−2`x`.

Find the $y$

`y`-value of the $y$`y`-intercept of the line.Find the $x$

`x`-value of the $x$`x`-intercept of the line.Find the value of $y$

`y`when $x=2$`x`=2.Plot the equation of the line below.

Loading Graph...

Consider the function $y=-\frac{1}{2}x^2$`y`=−12`x`2

Complete the following table of values.

$x$ `x`$-2$−2 $-1$−1 $0$0 $1$1 $2$2 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the graph.

Loading Graph...

Consider the cubic function $y=4x^3-3$`y`=4`x`3−3

Is the cubic increasing or decreasing from left to right?

Increasing

ADecreasing

BIs the cubic more or less steep than the function $y=x^3$

`y`=`x`3 ?More steep

ALess steep

BWhat are the coordinates of the point of inflection of the function?

Inflection ($\editable{}$, $\editable{}$)

Plot the graph $y=4x^3-3$

`y`=4`x`3−3Loading Graph...

Consider the function $y=-\frac{1}{4x}$`y`=−14`x`

Complete the following table of values.

$x$ `x`$-3$−3 $-2$−2 $-1$−1 $1$1 $2$2 $3$3 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Sketch the graph.

Loading Graph...In which quadrants does the graph lie?

$1$1

A$2$2

B$3$3

C$4$4

D