Graphs and vertical translations

 

There are certain basic functions that you should familiarise yourself with. They have names, forms and graphs. A summary of six interesting functions is shown here:

The linear function (far left)

Look at the linear function on the far left. It has the form $y=mx+b$y=mx+b. As its name suggests, it is drawn as a straight line that intersects the $y$y - axis at $\left(0,b\right)$(0,b) and leans according to the value of the coefficient $m$m.

For example the function $f\left(x\right)=2x-1$f(x)=2x−1 changes any $x$x value into a $y$y value determined as twice the $x$x value minus $1$1. So an $x$x value of $0$0 is mapped to to a $y$y value of $-1$−1. This mapping from $0$0 to $-1$−1 is summarised as the point $\left(0,-1\right)$(0,−1).

Similarly, the $x$x value of $1$1 is in turn mapped to $1$1, the $x$x value of $2$2 is mapped to $3$3, and the $x$x value of $3$3 is mapped to $5$5, and so the graph of the function passes through the four points $\left(0,-1\right),\left(1,1\right),\left(2,3\right)$(0,−1),(1,1),(2,3) and $\left(3,5\right)$(3,5), and the linearity becomes clear.

Each extra unit increase in $x$x induces an increase of two units in $y$y. Thus the gradient of the line is $2$2, the coefficient of $x$x in the function's rule $f\left(x\right)=2x-1$f(x)=2x−1.   

The Parabola (2nd from left)  

The second function, given as $y=x^2$y=x2, takes x values and squares them.  This function is known as a quadratic. Non-zero numbers when squared become positive, and that explains a key feature of the graph. Moreover as the absolute value of $x$x becomes large, $x^2$x2 becomes very large and that's why the curve becomes steep on both sides of the axis of symmetry. This curve, known as a parabola, plays a significant role  in many naturally occurring  phenomena. 

The cubic (3rd from left)

The cubic function $y=x^3$y=x3 grows even quicker than the parabola, but negative numbers when cubed remain negative. This explains the curve's shape. It is said to have $180^\circ$180°  rotational symmetry about the origin. 

 

The Exponential function (4th from left)

The exponential function $y=a^x,a>1$y=ax,a>1 has the independent variable $x$x in the exponent. So for example $y=2^x$y=2x maps the seven x values $-3,-2,-1,0,1,2,3$−3,−2,−1,0,1,2,3 to $\frac{1}{8},\frac{1}{4},\frac{1}{2},0,2,4,8$18​,14​,12​,0,2,4,8 respectively. This is why it has the  characteristic monotonically rising shape. The function values are always positive and increase rapidly as x continues to increase.

The Rectangular Hyperbola (5th from left)

The positive function values of $y=\frac{1}{x}$y=1x​ are large when $x$x is small and positive, and small when $x$x is large and positive. In other words, the function takes small values and turns them into large values and vice-versa. This is why the graph looks as it does.

The same is true for negative values, and this is why the hyperbola is another function that possesses $180^\circ$180°  rotational symmetry about the origin.    

The absolute value function

This function, given as $y=\left|x\right|$y=|x| leaves positive values of the line $y=x$y=x unchanged but changes the negative values of the same line to positive values. Negative values are thus reflected in the $x$x-axis, and this is why the function exhibits its angular shape.

Simple translations of the 6 graphs

Any of the above graphs can be moved up or down $k$k units by simply adding $k$k to the function. For example the graph of the function $y=x^2+3$y=x2+3 is essentially the same as $y=x^2$y=x2 but moved up (translated vertically) $3$3 units. The same idea applies to any of the six basic graphs.

Any of the graphs can be moved sideways $h$h units by simply replacing $x$x with $\left(x-h\right)$(x−h). So for example the cubic graph $y=\left(x-3\right)^3$y=(x−3)3 is essentially the same as $y=x^3$y=x3 but moved to the right (translated horizontally to the right) by $3$3 units. Again, The same idea applies to any of the six basic graphs.

As an interesting example let's graph the two functions $y=\frac{1}{x-3}+2$y=1x−3​+2 and $y=\left(x-1\right)^2$y=(x−1)2 on the same axes and see where they intersect.

The first function represents a rectangular hyperbola shifted $3$3 units to the right, and then lifted $2$2 units vertically upward. The second function represents a parabola shifted $1$1 unit to the right. 

The two graphs are shown here, intersecting in three places: 

Finding the intersections (extension only)

From the graph the two functions intersect at approximately $\left(-0.3,1.7\right),\left(2,1\right)$(−0.3,1.7),(2,1) and $\left(3.3,5.3\right)$(3.3,5.3). However, to find the exact intersections requires much more mathematics. If you are interested, you can try to follow the logic in the next paragraph. 

We equate the two functions so that $\left(x-1\right)^2=\frac{1}{x-3}+2$(x−1)2=1x−3​+2 and try to solve for $x$x. This leads us to the cubic equation $x^3-5x^2+5x+2=0$x3−5x2+5x+2=0. From the graph, or by testing directly into the cubic equation, we can see that $x=2$x=2 is a solution.

A neat piece of mathematics involves rearranging the cubic to $x^3-2x^2-3x^2+6x-x+2=0$x3−2x2−3x2+6x−x+2=0. Check this!

Thus, by factoring in pairs we have $x^2\left(x-2\right)-3x\left(x-2\right)-\left(x-2\right)=0$x2(x−2)−3x(x−2)−(x−2)=0, or when the common factor is taken out, $\left(x-2\right)\left(x^2-3x-1\right)=0$(x−2)(x2−3x−1)=0. 

We now can solve the quadratic equation $x^2-3x-1=0$x2−3x−1=0 using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b2−4ac2a​.

The other two $x$x values are thus given by $x=\frac{3\pm\sqrt{13}}{2}$x=3±√132​. Our estimate of $3.3$3.3 and $-0.3$−0.3 turn out to be very close. The $y$y values can be found by substituting these values into either equation.  

 

Worked Examples

Question 1

Question 2

Question 3