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Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of functions.

Functions of the Form y=ax+p+qy=ax+p+q

This form of the hyperbolic function is slightly more complex than the form studied in Grade 10.

Figure 1: General shape and position of the graph of a function of the form f(x)=ax+p+qf(x)=ax+p+q. The asymptotes are shown as dashed lines.
Figure 1 (MG11C12_001.png)

Investigation : Functions of the Form y=ax+p+qy=ax+p+q

  1. On the same set of axes, plot the following graphs:
    1. a(x)=-2x+1+1a(x)=-2x+1+1
    2. b(x)=-1x+1+1b(x)=-1x+1+1
    3. c(x)=0x+1+1c(x)=0x+1+1
    4. d(x)=1x+1+1d(x)=1x+1+1
    5. e(x)=2x+1+1e(x)=2x+1+1
    Use your results to deduce the effect of aa.
  2. On the same set of axes, plot the following graphs:
    1. f(x)=1x-2+1f(x)=1x-2+1
    2. g(x)=1x-1+1g(x)=1x-1+1
    3. h(x)=1x+0+1h(x)=1x+0+1
    4. j(x)=1x+1+1j(x)=1x+1+1
    5. k(x)=1x+2+1k(x)=1x+2+1
    Use your results to deduce the effect of pp.
  3. Following the general method of the above activities, choose your own values of aa and pp to plot 5 different graphs of y=ax+p+qy=ax+p+q to deduce the effect of qq.

You should have found that the sign of aa affects whether the graph is located in the first and third quadrants, or the second and fourth quadrants of Cartesian plane.

You should have also found that the value of pp affects whether the xx-intercept is negative (p>0p>0) or positive (p<0p<0).

You should have also found that the value of qq affects whether the graph lies above the xx-axis (q>0q>0) or below the xx-axis (q<0q<0).

These different properties are summarised in Table 1. The axes of symmetry for each graph is shown as a dashed line.

Table 1: Table summarising general shapes and positions of functions of the form y=ax+p+qy=ax+p+q. The axes of symmetry are shown as dashed lines.
  p < 0 p < 0 p > 0 p > 0
  a > 0 a > 0 a < 0 a < 0 a > 0 a > 0 a < 0 a < 0
q > 0 q > 0
Figure 2
Figure 2 (MG11C12_002.png)
Figure 3
Figure 3 (MG11C12_003.png)
Figure 4
Figure 4 (MG11C12_004.png)
Figure 5
Figure 5 (MG11C12_005.png)
q < 0 q < 0
Figure 6
Figure 6 (MG11C12_006.png)
Figure 7
Figure 7 (MG11C12_007.png)
Figure 8
Figure 8 (MG11C12_008.png)
Figure 9
Figure 9 (MG11C12_009.png)

Domain and Range

For y=ax+p+qy=ax+p+q, the function is undefined for x=-px=-p. The domain is therefore {x:x∈R,x≠-p}{x:x∈R,x≠-p}.

We see that y=ax+p+qy=ax+p+q can be re-written as:

y = a x + p + q y - q = a x + p If x ≠ - p then : ( y - q ) ( x + p ) = a x + p = a y - q y = a x + p + q y - q = a x + p If x ≠ - p then : ( y - q ) ( x + p ) = a x + p = a y - q
(1)

This shows that the function is undefined at y=qy=q. Therefore the range of f(x)=ax+p+qf(x)=ax+p+q is {f(x):f(x)∈R,f(x)≠q{f(x):f(x)∈R,f(x)≠q.

For example, the domain of g(x)=2x+1+2g(x)=2x+1+2 is {x:x∈R,x≠-1}{x:x∈R,x≠-1} because g(x)g(x) is undefined at x=-1x=-1.

y = 2 x + 1 + 2 ( y - 2 ) = 2 x + 1 ( y - 2 ) ( x + 1 ) = 2 ( x + 1 ) = 2 y - 2 y = 2 x + 1 + 2 ( y - 2 ) = 2 x + 1 ( y - 2 ) ( x + 1 ) = 2 ( x + 1 ) = 2 y - 2
(2)

We see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)∈(-∞,2)∪(2,∞)}{g(x):g(x)∈(-∞,2)∪(2,∞)}.

Domain and Range

  1. Determine the range of y=1x+1y=1x+1.
  2. Given:f(x)=8x-8+4f(x)=8x-8+4. Write down the domain of ff.
  3. Determine the domain of y=-8x+1+3y=-8x+1+3

Intercepts

For functions of the form, y=ax+p+qy=ax+p+q, the intercepts with the xx and yy axis are calculated by setting x=0x=0 for the yy-intercept and by setting y=0y=0 for the xx-intercept.

The yy-intercept is calculated as follows:

y = a x + p + q y i n t = a 0 + p + q = a p + q y = a x + p + q y i n t = a 0 + p + q = a p + q
(3)

For example, the yy-intercept of g(x)=2x+1+2g(x)=2x+1+2 is given by setting x=0x=0 to get:

y = 2 x + 1 + 2 y i n t = 2 0 + 1 + 2 = 2 1 + 2 = 2 + 2 = 4 y = 2 x + 1 + 2 y i n t = 2 0 + 1 + 2 = 2 1 + 2 = 2 + 2 = 4
(4)

The xx-intercepts are calculated by setting y=0y=0 as follows:

y = a x + p + q 0 = a x i n t + p + q a x i n t + p = - q a = - q ( x i n t + p ) x i n t + p = a - q x i n t = a - q - p y = a x + p + q 0 = a x i n t + p + q a x i n t + p = - q a = - q ( x i n t + p ) x i n t + p = a - q x i n t = a - q - p
(5)

For example, the xx-intercept of g(x)=2x+1+2g(x)=2x+1+2 is given by setting x=0x=0 to get:

y = 2 x + 1 + 2 0 = 2 x i n t + 1 + 2 - 2 = 2 x i n t + 1 - 2 ( x i n t + 1 ) = 2 x i n t + 1 = 2 - 2 x i n t = - 1 - 1 x i n t = - 2 y = 2 x + 1 + 2 0 = 2 x i n t + 1 + 2 - 2 = 2 x i n t + 1 - 2 ( x i n t + 1 ) = 2 x i n t + 1 = 2 - 2 x i n t = - 1 - 1 x i n t = - 2
(6)

Intercepts

  1. Given:h(x)=1x+4-2h(x)=1x+4-2. Determine the coordinates of the intercepts of hh with the x- and y-axes.
  2. Determine the x-intercept of the graph of y=5x+2y=5x+2. Give the reason why there is no y-intercept for this function.

Asymptotes

There are two asymptotes for functions of the form y=ax+p+qy=ax+p+q. They are determined by examining the domain and range.

We saw that the function was undefined at x=-px=-p and for y=qy=q. Therefore the asymptotes are x=-px=-p and y=qy=q.

For example, the domain of g(x)=2x+1+2g(x)=2x+1+2 is {x:x∈R,x≠-1}{x:x∈R,x≠-1} because g(x)g(x) is undefined at x=-1x=-1. We also see that g(x)g(x) is undefined at y=2y=2. Therefore the range is {g(x):g(x)∈(-∞,2)∪(2,∞)}{g(x):g(x)∈(-∞,2)∪(2,∞)}.

From this we deduce that the asymptotes are at x=-1x=-1 and y=2y=2.

Asymptotes

  1. Given:h(x)=1x+4-2h(x)=1x+4-2.Determine the equations of the asymptotes of hh.
  2. Write down the equation of the vertical asymptote of the graph y=1x-1y=1x-1.

Sketching Graphs of the Form f(x)=ax+p+qf(x)=ax+p+q

In order to sketch graphs of functions of the form, f(x)=ax+p+qf(x)=ax+p+q, we need to calculate four characteristics:

  1. domain and range
  2. asymptotes
  3. yy-intercept
  4. xx-intercept

For example, sketch the graph of g(x)=2x+1+2g(x)=2x+1+2. Mark the intercepts and asymptotes.

We have determined the domain to be {x:x∈R,x≠-1}{x:x∈R,x≠-1} and the range to be {g(x):g(x)∈(-∞,2)∪(2,∞)}{g(x):g(x)∈(-∞,2)∪(2,∞)}. Therefore the asymptotes are at x=-1x=-1 and y=2y=2.

The yy-intercept is yint=4yint=4 and the xx-intercept is xint=-2xint=-2.

Figure 10: Graph of g(x)=2x+1+2g(x)=2x+1+2.
Figure 10 (MG11C12_010.png)

Graphs

  1. Draw the graph of y=1x+2y=1x+2. Indicate the horizontal asymptote.
  2. Given:h(x)=1x+4-2h(x)=1x+4-2. Sketch the graph of hh showing clearly the asymptotes and ALL intercepts with the axes.
  3. Draw the graph of y=1xy=1x and y=-8x+1+3y=-8x+1+3 on the same system of axes.
  4. Draw the graph of y=5x-2,5+2y=5x-2,5+2. Explain your method.
  5. Draw the graph of the function defined by y=8x-8+4y=8x-8+4. Indicate the asymptotes and intercepts with the axes.

End of Chapter Exercises

  1. Plot the graph of the hyperbola defined by y=2xy=2x for -4≤x≤4-4≤x≤4. Suppose the hyperbola is shifted 3 units to the right and 1 unit down. What is the new equation then ?
  2. Based on the graph of y=1xy=1x, determine the equation of the graph with asymptotes y=2y=2 and x=1x=1 and passing through the point (2; 3).

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