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The parabola

Functions of the Form y=ax2+qy=ax2+q

The general shape and position of the graph of the function of the form f(x)=ax2+qf(x)=ax2+q, called a parabola, is shown in Figure 1. These are parabolic functions.

Investigation : Functions of the Form y=ax2+qy=ax2+q

1. On the same set of axes, plot the following graphs:
1. a(x)=-2·x2+1a(x)=-2·x2+1
2. b(x)=-1·x2+1b(x)=-1·x2+1
3. c(x)=0·x2+1c(x)=0·x2+1
4. d(x)=1·x2+1d(x)=1·x2+1
5. e(x)=2·x2+1e(x)=2·x2+1
Use your results to deduce the effect of aa.
2. On the same set of axes, plot the following graphs:
1. f(x)=x2-2f(x)=x2-2
2. g(x)=x2-1g(x)=x2-1
3. h(x)=x2+0h(x)=x2+0
4. j(x)=x2+1j(x)=x2+1
5. k(x)=x2+2k(x)=x2+2
Use your results to deduce the effect of qq.

The effects of aa and qq are a few examples are what are collectively called transformations of the function y=x2y=x2. The effect of qq is what is called a translation because all points are moved the same distance in the same direction, specifically a vertical translation (because it slides the graph up and down) of x2x2 by qq (note that a horizontal translation by qq would look like y=(x-q)2y=(x-q)2). The effect of aa is what is called scaling. When a>1a>1, it is usually called a stretch because the effect on the graph is to stretch it apart, and when a<1a<1 (but greater than zero), it is often called a shrink because it makes the graph shrink in on itself. When a= -1a=-1, it is called a reflection, specifically a reflection about the xx-axis because the xx-axis acts as a mirror for the graph. If aa is some other negative number, one would refer to it as a scaled reflection (or, alternatively, a 'stretched' or 'shrunk' reflection), because it is reflected and scaled. One often refers to reflections, translations, scaling, etc., as the effect of the transformation on the given function.

A more in-depth look at transformations can be found here. This section on transformations is simply an extension and is not needed for examinations.

Complete the following table of values for the functions aa to kk to help with drawing the required graphs in this activity:

 x x - 2 - 2 - 1 - 1 0 0 1 1 2 2 a ( x ) a ( x ) b ( x ) b ( x ) c ( x ) c ( x ) d ( x ) d ( x ) e ( x ) e ( x ) f ( x ) f ( x ) g ( x ) g ( x ) h ( x ) h ( x ) j ( x ) j ( x ) k ( x ) k ( x )

This simulation allows you to visualise the effect of changing a and q. Note that in this simulation q = c. Also an extra term bx has been added in. You can leave bx as 0, or you can also see what effect this has on the graph.

Figure 2
Phet simulation for graphing

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 is shown in Figure 3.

You should have also found that the value of qq affects whether the turning point is to the left of the yy-axis (q>0q>0) or to the right of the yy-axis (q<0q<0).

These different properties are summarised in Table 2.

 a > 0 a > 0 a < 0 a < 0 q > 0 q > 0 q < 0 q < 0

Domain and Range

For f(x)=ax2+qf(x)=ax2+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)=ax2+qf(x)=ax2+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 2 0 ( The square of an expression is always positive ) a x 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a x 2 + q q f ( x ) q x 2 0 ( The square of an expression is always positive ) a x 2 0 ( Multiplication by a positive number maintains the nature of the inequality ) a x 2 + q q f ( x ) q
(1)

This tells us that for all values of xx, f(x)f(x) is always greater than qq. Therefore if a>0a>0, the range of f(x)=ax2+qf(x)=ax2+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)=ax2+qf(x)=ax2+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)=x2+2g(x)=x2+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 2 0 x 2 + 2 2 g ( x ) 2 x 2 0 x 2 + 2 2 g ( x ) 2
(2)

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

Intercepts

For functions of the form, y=ax2+qy=ax2+q, the details of calculating the intercepts with the xx and yy axis is given.

The yy-intercept is calculated as follows:

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

For example, the yy-intercept of g(x)=x2+2g(x)=x2+2 is given by setting x=0x=0 to get:

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

The xx-intercepts are calculated as follows:

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

However, Equation 5 is only valid if -qa0-qa0 which means that either q0q0 or a<0a<0. 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 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.

If q=0q=0 then we have one intercept at x=0x=0.

For example, the xx-intercepts of g(x)=x2+2g(x)=x2+2 is given by setting y=0y=0 to get:

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

which is not real. Therefore, the graph of g(x)=x2+2g(x)=x2+2 does not have any xx-intercepts.

Turning Points

The turning point of the function of the form f(x)=ax2+qf(x)=ax2+q is given by examining the range of the function. We know that if a>0a>0 then the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)[q,)}{f(x):f(x)[q,)} and if a<0a<0 then the range of f(x)=ax2+qf(x)=ax2+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 t p 2 + q 0 = a x t p 2 0 = x t p 2 x t p = 0 q = a x t p 2 + q 0 = a x t p 2 0 = x t p 2 x t p = 0
(7)

x=0x=0 at f(x)=qf(x)=q. The co-ordinates of the (minimal) turning point is therefore (0,q)(0,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 (0,q)(0,q).

Axes of Symmetry

There is one axis of symmetry for the function of the form f(x)=ax2+qf(x)=ax2+q that passes through the turning point. Since the turning point lies on the yy-axis, the axis of symmetry is the yy-axis.

Sketching Graphs of the Form f(x)=ax2+qf(x)=ax2+q

In order to sketch graphs of the form, f(x)=ax2+qf(x)=ax2+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)=-12x2-3g(x)=-12x2-3. 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 2 0 - 1 2 x 2 0 - 1 2 x 2 - 3 - 3 f ( x ) - 3 x 2 0 - 1 2 x 2 0 - 1 2 x 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 2 - 3 = - 3 - 1 2 x 2 - 3 + 3 = 0 - 1 2 x 2 = 0 Divide both sides by - 1 2 : x 2 = 0 Take square root of both sides: x = 0 x = 0 - 1 2 x 2 - 3 = - 3 - 1 2 x 2 - 3 + 3 = 0 - 1 2 x 2 = 0 Divide both sides by - 1 2 : x 2 = 0 Take square root of both sides: x = 0 x = 0
(9)

The coordinates of the turning point are: (0;-3)(0;-3).

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

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

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

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

which is not real. Therefore, there are no xx-intercepts which means that the function does not cross or even touch the xx-axis at any point.

We also know that the axis of symmetry is the yy-axis.

Finally, we draw the graph. Note that in the diagram only the y-intercept is marked. The graph has a maximal turning point (i.e. makes a frown) as determined from the sign of a, there are no x-intercepts and the turning point is that same as the y-intercept. The domain is all real numbers and the range is {f(x):f(x)(-,-3]}{f(x):f(x)(-,-3]}.

Exercise 1: Drawing a parabola

Draw the graph of y=3x2+5y=3x2+5.

Solution
1. Step 1. Determine the sign of aa: The sign of aa is positive. The parabola will therefore have a minimal turning point.
2. Step 2. Find the domain and range: The domain is: {x:xR}{x:xR} and the range is: {f(x):f(x)[5,)}{f(x):f(x)[5,)}.
3. Step 3. Find the turning point: The turning point occurs at (0,q)(0,q). For this function q=5q=5, so the turning point is at (0,5)(0,5)
4. Step 4. Calculate the y-intercept: The y-intercept occurs when x=0x=0. Calculating the y-intercept gives:
y = 3x2+5 yint = 3(0)2+5 yint = 5 y = 3x2+5 yint = 3(0)2+5 yint = 5
(12)
5. Step 5. Calculate the x-intercept: The x-intercepts occur when y=0y=0. Calculating the x-intercept gives:
y = 3x2+5 0 = 3x2+5 x2 = -35 y = 3x2+5 0 = 3x2+5 x2 = -35
(13)
which is not real, so there are no x-intercepts.
6. Step 6. Plot the graph: Putting all this together gives us the following graph:

The following video shows one method of graphing parabolas. Note that in this video the term vertex is used in place of turning point. The vertex and the turning point are the same thing.

Figure 10
Khan academy video on graphing parabolas - 1

Parabolas

1. Show that if a<0a<0 that the range of f(x)=ax2+qf(x)=ax2+q is {f(x):f(x)(-;q]}{f(x):f(x)(-;q]}. Click here for the solution
2. Draw the graph of the function y=-x2+4y=-x2+4 showing all intercepts with the axes. Click here for the solution
3. Two parabolas are drawn: g:y=ax2+pg:y=ax2+p and h:y=bx2+qh:y=bx2+q.
1. Find the values of aa and pp.
2. Find the values of bb and qq.
3. Find the values of xx for which ax2+pbx2+qax2+pbx2+q.
4. For what values of xx is gg increasing ?

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