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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 quadratic functions.

Functions of the Form y=a(x+p)2+qy=a(x+p)2+q

This form of the quadratic function is slightly more complex than the form studied in Grade 10, y=ax2+qy=ax2+q. The general shape and position of the graph of the function of the form f(x)=a(x+p)2+qf(x)=a(x+p)2+q is shown in Figure 1.

Figure 1: Graph of f(x)=12(x+2)2-1f(x)=12(x+2)2-1
Figure 1 (MG11C11_001.png)

Investigation : Functions of the Form y=a(x+p)2+qy=a(x+p)2+q

  1. On the same set of axes, plot the following graphs:
    1. a(x)=(x-2)2a(x)=(x-2)2
    2. b(x)=(x-1)2b(x)=(x-1)2
    3. c(x)=x2c(x)=x2
    4. d(x)=(x+1)2d(x)=(x+1)2
    5. e(x)=(x+2)2e(x)=(x+2)2
    Use your results to deduce the effect of pp.
  2. On the same set of axes, plot the following graphs:
    1. f(x)=(x-2)2+1f(x)=(x-2)2+1
    2. g(x)=(x-1)2+1g(x)=(x-1)2+1
    3. h(x)=x2+1h(x)=x2+1
    4. j(x)=(x+1)2+1j(x)=(x+1)2+1
    5. k(x)=(x+2)2+1k(x)=(x+2)2+1
    Use your results to deduce the effect of qq.
  3. Following the general method of the above activities, choose your own values of pp and qq to plot 5 different graphs (on the same set of axes) of y=a(x+p)2+qy=a(x+p)2+q to deduce the effect of aa.

From your graphs, you should have found that aa affects whether the graph makes a smile or a frown. If a<0a<0, the graph makes a frown and if a>0a>0 then the graph makes a smile. This was shown in Grade 10.

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

You should have also found that the value of pp affects whether the turning point is to the left of the yy-axis (p>0p>0) or to the right of the yy-axis (p<0p<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=a(x+p)2+qy=a(x+p)2+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 (MG11C11_002.png)
Figure 3
Figure 3 (MG11C11_003.png)
Figure 4
Figure 4 (MG11C11_004.png)
Figure 5
Figure 5 (MG11C11_005.png)
q 0 q 0
Figure 6
Figure 6 (MG11C11_006.png)
Figure 7
Figure 7 (MG11C11_007.png)
Figure 8
Figure 8 (MG11C11_008.png)
Figure 9
Figure 9 (MG11C11_009.png)

Figure 10
Phet simulation for graphing

Domain and Range

For f(x)=a(x+p)2+qf(x)=a(x+p)2+q, the domain is {x:xR}{x:xR} because there is no value of xRxR for which f(x)f(x) is undefined.

The range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q depends on whether the value for aa is positive or negative. We will consider these two cases separately.

If a>0a>0 then we have:

( x + p ) 2 0 ( The square of an expression is always positive ) a ( x + p ) 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a ( x + p ) 2 + q q f ( x ) q ( x + p ) 2 0 ( The square of an expression is always positive ) a ( x + p ) 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a ( x + p ) 2 + q q f ( x ) q
(1)

This tells us that for all values of xx, f(x)f(x) is always greater than or equal to qq. Therefore if a>0a>0, the range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q is {f(x):f(x)[q,)}{f(x):f(x)[q,)}.

Similarly, it can be shown that if a<0a<0 that the range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q is {f(x):f(x)(-,q]}{f(x):f(x)(-,q]}. This is left as an exercise.

For example, the domain of g(x)=(x-1)2+2g(x)=(x-1)2+2 is {x:xR}{x:xR} because there is no value of xRxR for which g(x)g(x) is undefined. The range of g(x)g(x) can be calculated as follows:

( x - p ) 2 0 ( x + p ) 2 + 2 2 g ( x ) 2 ( x - p ) 2 0 ( x + p ) 2 + 2 2 g ( x ) 2
(2)

Therefore the range is {g(x):g(x)[2,)}{g(x):g(x)[2,)}.

Domain and Range

  1. Given the function f(x)=(x-4)2-1f(x)=(x-4)2-1. Give the range of f(x)f(x).
  2. What is the domain of the equation y=2x2-5x-18y=2x2-5x-18 ?

Intercepts

For functions of the form, y=a(x+p)2+qy=a(x+p)2+q, the details of calculating the intercepts with the xx and yy axes is given.

The yy-intercept is calculated as follows:

y = a ( x + p ) 2 + q y i n t = a ( 0 + p ) 2 + q = a p 2 + q y = a ( x + p ) 2 + q y i n t = a ( 0 + p ) 2 + q = a p 2 + q
(3)

If p=0p=0, then yint=qyint=q.

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

g ( x ) = ( x - 1 ) 2 + 2 y i n t = ( 0 - 1 ) 2 + 2 = ( - 1 ) 2 + 2 = 1 + 2 = 3 g ( x ) = ( x - 1 ) 2 + 2 y i n t = ( 0 - 1 ) 2 + 2 = ( - 1 ) 2 + 2 = 1 + 2 = 3
(4)

The xx-intercepts are calculated as follows:

y = a ( x + p ) 2 + q 0 = a ( x i n t + p ) 2 + q a ( x i n t + p ) 2 = - q x i n t + p = ± - q a x i n t = ± - q a - p y = a ( x + p ) 2 + q 0 = a ( x i n t + p ) 2 + q a ( x i n t + p ) 2 = - q x i n t + p = ± - q a x i n t = ± - q a - p
(5)

However, Equation 5 is only valid if -qa>0-qa>0 which means that either q<0q<0 or a<0a<0 but not both. This is consistent with what we expect, since if q>0q>0 and a>0a>0 then -qa-qa is negative and in this case the graph lies above the xx-axis and therefore does not intersect the xx-axis. If however, q>0q>0 and a<0a<0, then -qa-qa is positive and the graph is hat shaped with turning point above the xx-axis and should have two xx-intercepts. Similarly, if q<0q<0 and a>0a>0 then -qa-qa is also positive, and the graph should intersect with the xx-axis twice.

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

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

which has no real solutions. Therefore, the graph of g(x)=(x-1)2+2g(x)=(x-1)2+2 does not have any xx-intercepts.

Intercepts

  1. Find the x- and y-intercepts of the function f(x)=(x-4)2-1f(x)=(x-4)2-1.
  2. Find the intercepts with both axes of the graph of f(x)=x2-6x+8f(x)=x2-6x+8.
  3. Given: f(x)=-x2+4x-3f(x)=-x2+4x-3. Calculate the x- and y-intercepts of the graph of ff.

Turning Points

The turning point of the function of the form f(x)=a(x+p)2+qf(x)=a(x+p)2+q is given by examining the range of the function. We know that if a>0a>0 then the range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q is {f(x):f(x)[q,)}{f(x):f(x)[q,)} and if a<0a<0 then the range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q is {f(x):f(x)(-,q]}{f(x):f(x)(-,q]}.

So, if a>0a>0, then the lowest value that f(x)f(x) can take on is qq. Solving for the value of xx at which f(x)=qf(x)=q gives:

q = a ( x + p ) 2 + q 0 = a ( x + p ) 2 0 = ( x + p ) 2 0 = x + p x = - p q = a ( x + p ) 2 + q 0 = a ( x + p ) 2 0 = ( x + p ) 2 0 = x + p x = - p
(7)

x=-px=-p at f(x)=qf(x)=q. The co-ordinates of the (minimal) turning point is therefore (-p,q)(-p,q).

Similarly, if a<0a<0, then the highest value that f(x)f(x) can take on is qq and the co-ordinates of the (maximal) turning point is (-p,q)(-p,q).

Turning Points

  1. Determine the turning point of the graph of f(x)=x2-6x+8f(x)=x2-6x+8 .
  2. Given: f(x)=-x2+4x-3f(x)=-x2+4x-3. Calculate the co-ordinates of the turning point of ff.
  3. Find the turning point of the following function by completing the square: y=12(x+2)2-1y=12(x+2)2-1.

Axes of Symmetry

There is only one axis of symmetry for the function of the form f(x)=a(x+p)2+qf(x)=a(x+p)2+q. This axis passes through the turning point and is parallel to the yy-axis. Since the xx-coordinate of the turning point is x=-px=-p, this is the axis of symmetry.

Axes of Symmetry

  1. Find the equation of the axis of symmetry of the graph y=2x2-5x-18y=2x2-5x-18.
  2. Write down the equation of the axis of symmetry of the graph of y=3(x-2)2+1y=3(x-2)2+1.
  3. Write down an example of an equation of a parabola where the y-axis is the axis of symmetry.

Sketching Graphs of the Form f(x)=a(x+p)2+qf(x)=a(x+p)2+q

In order to sketch graphs of the form f(x)=a(x+p)2+qf(x)=a(x+p)2+q, we need to determine five characteristics:

  1. sign of aa
  2. domain and range
  3. turning point
  4. yy-intercept
  5. xx-intercept

For example, sketch the graph of g(x)=-12(x+1)2-3g(x)=-12(x+1)2-3. Mark the intercepts, turning point and axis of symmetry.

Firstly, we determine that a<0a<0. This means that the graph will have a maximal turning point.

The domain of the graph is {x:xR}{x:xR} because f(x)f(x) is defined for all xRxR. The range of the graph is determined as follows:

( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 - 3 - 3 f ( x ) - 3 ( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 0 - 1 2 ( x + 1 ) 2 - 3 - 3 f ( x ) - 3
(8)

Therefore the range of the graph is {f(x):f(x)(-,-3]}{f(x):f(x)(-,-3]}.

Using the fact that the maximum value that f(x)f(x) achieves is -3, then the yy-coordinate of the turning point is -3. The xx-coordinate is determined as follows:

- 1 2 ( x + 1 ) 2 - 3 = - 3 - 1 2 ( x + 1 ) 2 - 3 + 3 = 0 - 1 2 ( x + 1 ) 2 = 0 Divide both sides by - 1 2 : ( x + 1 ) 2 = 0 Take square root of both sides: x + 1 = 0 x = - 1 - 1 2 ( x + 1 ) 2 - 3 = - 3 - 1 2 ( x + 1 ) 2 - 3 + 3 = 0 - 1 2 ( x + 1 ) 2 = 0 Divide both sides by - 1 2 : ( x + 1 ) 2 = 0 Take square root of both sides: x + 1 = 0 x = - 1
(9)

The coordinates of the turning point are: (-1,-3)(-1,-3).

The yy-intercept is obtained by setting x=0x=0. This gives:

y i n t = - 1 2 ( 0 + 1 ) 2 - 3 = - 1 2 ( 1 ) - 3 = - 1 2 - 3 = - 1 2 - 3 = - 7 2 y i n t = - 1 2 ( 0 + 1 ) 2 - 3 = - 1 2 ( 1 ) - 3 = - 1 2 - 3 = - 1 2 - 3 = - 7 2
(10)

The xx-intercept is obtained by setting y=0y=0. This gives:

0 = - 1 2 ( x i n t + 1 ) 2 - 3 3 = - 1 2 ( x i n t + 1 ) 2 - 3 · 2 = ( x i n t + 1 ) 2 - 6 = ( x i n t + 1 ) 2 0 = - 1 2 ( x i n t + 1 ) 2 - 3 3 = - 1 2 ( x i n t + 1 ) 2 - 3 · 2 = ( x i n t + 1 ) 2 - 6 = ( x i n t + 1 ) 2
(11)

which has no real solutions. Therefore, there are no xx-intercepts.

We also know that the axis of symmetry is parallel to the yy-axis and passes through the turning point.

Figure 11: Graph of the function f(x)=-12(x+1)2-3f(x)=-12(x+1)2-3
Figure 11 (MG11C11_010.png)

Figure 12
Khan academy video on graphing quadratics

Sketching the Parabola

  1. Draw the graph of y=3(x-2)2+1y=3(x-2)2+1 showing all the intercepts with the axes as well as the coordinates of the turning point.
  2. Draw a neat sketch graph of the function defined by y=ax2+bx+cy=ax2+bx+c if a>0a>0; b<0b<0; b2=4acb2=4ac.

Writing an equation of a shifted parabola

Given a parabola with equation y=x2-2x-3y=x2-2x-3. The graph of the parabola is shifted one unit to the right. Or else the y-axis shifts one unit to the left i.e. xx becomes x-1x-1. Therefore the new equation will become:

y = ( x - 1 ) 2 - 2 ( x - 1 ) - 3 = x 2 - 2 x + 1 - 2 x + 2 - 3 = x 2 - 4 x y = ( x - 1 ) 2 - 2 ( x - 1 ) - 3 = x 2 - 2 x + 1 - 2 x + 2 - 3 = x 2 - 4 x
(12)

If the given parabola is shifted 3 units down i.e. yy becomes y+3y+3. The new equation will be:

(Notice the x-axis then moves 3 units upwards)

y + 3 = x 2 - 2 x - 3 y = x 2 - 2 x - 6 y + 3 = x 2 - 2 x - 3 y = x 2 - 2 x - 6
(13)

End of Chapter Exercises

  1. Show that if a<0a<0, then the range of f(x)=a(x+p)2+qf(x)=a(x+p)2+q is {f(x):f(x)(-,q]}{f(x):f(x)(-,q]}.
  2. If (2,7) is the turning point of f(x)=-2x2-4ax+kf(x)=-2x2-4ax+k, find the values of the constants aa and kk.
  3. The graph in the figure is represented by the equation f(x)=ax2+bxf(x)=ax2+bx. The coordinates of the turning point are (3,9). Show that a=-1a=-1 and b=6b=6.
    Figure 13
    Figure 13 (MG11C11_011.png)
  4. Given: y=x2-2x+3y=x2-2x+3. Give the equation of the new graph originating if:
    1. The graph of ff is moved three units to the left.
    2. The xx-axis is moved down three units.
  5. A parabola with turning point (-1,-4) is shifted vertically by 4 units upwards. What are the coordinates of the turning point of the shifted parabola?

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