Mixed graphs II (linear, quad, cubic, 1/x)

Reviewing some basic graph types

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=mx+b is shown on the left and the hyperbola $y=\frac{1}{x}$y=1x​ is shown second from right. 

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

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

Translations

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=1x​ 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=1x−3​, or expressed explicitly $y=\frac{1}{x-3}-5$y=1x−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=x2 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=x3 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. 

Dilations and reflections

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=x2 to $y=3x^2$y=3x2 is the dilation factor $3$3. Every function value in $y=3x^2$y=3x2 is $3$3 times the associated function value in $y=x^2$y=x2. This means that the curve becomes steeper.

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

Every function value of $y=3\left(2^x\right)$y=3(2x) is $3$3 times that of $y=2^x$y=2x, and every function value of $y=\frac{1}{2}x^3$y=12​x3 is one-halve that of $y=x^3$y=x3.

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=−x2 is the negative of those of $y=x^2$y=x2, and the curve becomes an upside-down parabola with the same vertex as $y=x^2$y=x2. Again, every function value of $y=-\frac{3}{x}$y=−3x​ is the reflected image of across the x -axis of $y=\frac{3}{x}$y=3x​.  

 

Combining Translations and dilations

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=5x−2​+3 is constructed as follows. Start with the curve given by the basic function $y=\frac{1}{x}$y=1x​ 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=1x−2​+3. Note how changing from $y=\frac{1}{x-2}+3$y=1x−2​+3 to $y=\frac{5}{x-2}+3$y=5x−2​+3 pushes the hyperbola out diagonally. 

Finding $x$x and $y$y intercepts    

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=1x−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$1x−3​−5=0. This means that $\frac{1}{x-3}=5$1x−3​=5 and inverting $x-3=\frac{1}{5}$x−3=15​ so that $x=3\frac{1}{5}$x=315​.

 

 

 

Worked Examples

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Question 4