Introduction
Linear equations and methods for their solution were introduced in the previous
chapter. As we have seen, the graphs of linear functions are straight lines and
therefore their slopes are constant. This means that the function changes by a
constant amount whenever the dependent variable changes by the same fixed
value. This type of behaviour is not always observed in real-life applications in
economics. It is, therefore, necessary to introduce an added level of sophistication
to the mathematical modelling. This is achieved through the introduction
of nonlinear functions. The simplest nonlinear function is the quadratic function.
This function takes the general form
f(x) = ax2 + bx + c, (3.1)
where a = 0, b and c are constants. The condition a = 0 is to prevent the
occurrence of the degenerate case in which (3.1) reduces to a linear function.
If the profit function for a firm is given by a quadratic expression, then one
can determine the level of output for which the firm breaks even by solving a
quadratic equation. Additionally, one can determine the maximum profit and
the level of output for which it is attained by algebraically manipulating the
expression for the function. For more general nonlinear functions, the maximum
and/or minimum values of a function can be determined using the techniques
of calculus (see Chapter 7), but for a quadratic function this can be achieved
using algebra.
49
50 Elements of Mathematics for Economics and Finance
Certain total cost and total revenue functions are examples of quadratic
functions and are defined in terms of a quadratic expression involving the demand.
3.2 Graphs of Quadratic Functions
In the case of a linear function of the form f(x) = dx+e, the parameters d and
e can be interpreted in terms of properties of the graph of the function. The
value of d, the coefficient of x, gives the slope or gradient of the function, and
the value of e, the constant term, tells us where the straight line intercepts the
y-axis. A natural question to ask is whether the parameters in the expression
defining the general quadratic function f(x) = ax2 +bx+c can be interpreted
in a similar way in order to help us sketch its graph.
If we evaluate the function f(x) = ax2 + bx + c when x = 0 we obtain
f(0) = c. Therefore, the quadratic function intercepts the y-axis at the location
y = c. The values of the other parameters cannot be interpreted in such a
simple manner. However, the sign of the parameter a tells us something about
the shape of the graph. If a > 0, then the graph of f(x) has a

shape, whereas
if a < 0 the graph of f(x) has a shape. This information gives us a rough idea of what the graph of a quadratic function looks like. An additional aid is to tabulate the function at a sequence of integer values of x and to draw a smooth curve through the set of points. For example, let us sketch the graph of the quadratic function f(x) = x2 for −3 ≤ x ≤ 3. If we compare the coefficients of this function with those of the general quadratic function, we find that a = 1, and b = c = 0. Therefore, the graph of this function intercepts the y-axis at the origin as c = 0 and has a shape as a > 0. The values of this function are
tabulated in Table 3.1 for integer values of x for which −3 ≤ x ≤ 3, and the
graph of the function is shown in Fig. 3.1.
Now consider the function f(x) = 2×2 +3x−2. Again comparison with the
general quadratic function (3.1) shows that a = 2, b = 3, and c = −2. The
graph is again of a

shape since a > 0 and it intercepts the y-axis at y = −2.
The values of this function for integer values of x between −4 and 2 are shown
Table 3.1 Table of values of the function f(x) = x2 for integer values of
x for which −3 ≤ x ≤ 3. The graph of this function is shown in Fig. 3.1.
x −3 −2 −1 0 1 2 3
f(x) 9 4 1 0 1 4 9

3. Quadratic Equations 51
   x
   -3 -2 -1 0 1 2 3
   1
   2
   3
   4
   5
   6
   7
   8
   y 9
   Figure 3.1 The graph of the function f(x) = x2 for −3 ≤ x ≤ 3.
   in Table 3.2, and the graph of the function is shown in Fig. 3.2. The graph of
   this function crosses the x-axis in two places, at x = −2 and x = 1/2. These
   values of x satisfy the quadratic equation 2×2 + 3x −2 = 0 since y = f(x) = 0
   at these two points. The values of x that satisfy the equation f(x) = 0 are
   known as the roots or solutions of the equation. These two terms are used
   interchangeably. Therefore, we say that x = −2 and x = 1/2 are the roots or
   solutions of the quadratic equation 2×2 + 3x − 2 = 0.
   The next function we consider is f(x) = 2x − x2. This function has a
   Table 3.2 Table of values of the function f(x) = 2×2 +3x − 2 for integer
   values of x for which −4 ≤ x ≤ 2. The graph of this function is shown in Fig.
   3.2.
   x −4 −3 −2 −1 0 1 2
   2×2 32 18 8 2 0 2 8
   3x −12 −9 −6 −3 0 3 6
   −2 −2 −2 −2 −2 −2 −2 −2
   f(x) 18 7 0 −3 −2 3 12
   52 Elements of Mathematics for Economics and Finance
   x
   -4 -3 -2 -1 0 1 2
   5
   10
   15
   y
   Figure 3.2 The graph of the function f(x) = 2×2+3x−2 for −4 ≤ x ≤ 2.
   negative coefficient of x2. In terms of the general quadratic function (3.1), we
   have a = −1, b = 2, and c = 0. Since a < 0, the graph of the function has a
   
   shape. The graph intersects the x-axis at the origin since y = 2x − x2 = when
   x = 0. The other intercept (intersection) with the x-axis is the other root of the
   equation 2x − x2 = 0, namely x = 2. This is evident since 2x − x2 = (2 − x)x.
   Therefore, one of the roots of the equation 2x−x2 = 0 is x = 0. The other root
   is x = 2. The values of this function for integer values of x between −2 and 4
   are shown in Table 3.3, and the graph of the function is shown in Fig. 3.3.
   Finally, we consider the function f(x) = x2 − 2x + 2. Comparison with
   the general quadratic function (3.1) gives a = 1, b = −2, and c = 2. The
   Table 3.3 Table of values of the function f(x) = 2x−x2 for integer values
   of x for which −2 ≤ x ≤ 4. The graph of this function is shown in Fig. 3.3.
   x −2 −1 0 1 2 3 4
   2x −4 −2 0 2 4 6 8
   −x2 −4 −1 0 −1 −4 −9 −16
   f(x) −8 −3 0 1 0 −3 −8
4. Quadratic Equations 53
   x
   -2 -1 0 1 2 3 4
   -8
   -6
   -4
   -2
   2
   y
   Figure 3.3 The graph of the function f(x) = 2x − x2 for −2 ≤ x ≤ 4.
   values of this function at integer values of x between −2 and 4 are shown in
   Table 3.4, and the graph of the function is shown in Fig. 3.4. Note that the
   graph of this function does not cross the x-axis. It lies entirely above the xaxis,
   i.e., f(x) > 0 for all values of x. Therefore, there are no real roots of the
   corresponding equation x2 − 2x + 2 = 0.
   The graph of a quadratic function is known as a parabola. On inspection
   of Figs. 3.1–3.4, we observe that a parabola is symmetric about a vertical line
   x = h, where h is some constant. This line is known as the axis of symmetry of
   Table 3.4 Table of values of the function f(x) = x2 − 2x + 2 for integer
   values of x for which −2 ≤ x ≤ 4. The graph of this function is shown in Fig.
   3.4.
   x −2 −1 0 1 2 3 4
   x2 4 1 0 1 4 9 16
   −2x 4 2 0 −2 −4 −6 −8
   2 2 2 2 2 2 2 2
   f(x) 10 5 2 1 2 5 10
   54 Elements of Mathematics for Economics and Finance
   x
   -2 -1 0 1 2 3 4
   2
   4
   6
   8
   10
   y
   Figure 3.4 The graph of the function f(x) = x2 −2x+2 for −3 ≤ x ≤ 3.
   the parabola. The point of intersection of a parabola with its axis of symmetry
   is called the vertex. For example, the quadratic function f(x) = x2 − 3x + 2
   has x = 3/2 as its axis of symmetry and (3/2,−1/4) as its vertex. If a > 0, then
   the y component of the vertex provides the minimum value of the quadratic
   function. Similarly, if a < 0, then the y component of the vertex provides the
   maximum value of the quadratic function.
   If a quadratic function can be expressed in the form
   f(x) = a(x − h)2 + k, (3.2)
   then the axis of symmetry is x − h = 0 and the vertex is the point with coordinates
   (h, k). Let us rearrange the expression defining the general quadratic
   expression so that it is in this form. To do this, we use a process known as completing
   the square. First, we extract a factor a from the quadratic expression
   ax2 + bx + c, i.e.,
   ax2 + bx + c = a
   
   x2 + b
   a
   x + c
   a
   
   (3.3)
   Then, we express the first two terms inside the bracket on the right-hand side
   of (3.3), viz. x2 + (b/a)x as the difference between two squares:
   x2 + b
   a
   x =
   
   x + b
   2a
   2
   −
   
   b
   2a
   2
   .
5. Quadratic Equations 55
   Therefore,
   ax2 + bx + c = a
   
   x + b
   2a
   2
   − b2
   4a2 + c
   a
   
   (3.4)
   = a
   
   x + b
   2a
   2
   +
   4ac − b2
   4a2
   
   , (3.5)
   in which the last two terms in (3.4) have been combined to form a single
   fraction. Finally, we arrive at
   ax2 + bx + c = a
   
   x + b
   2a
   2
   +
   4ac − b2
   4a
   . (3.6)
   So comparing (3.6) with (3.2), we have
   h = − b
   2a
   , k =
   4ac − b2
   4a
   .
   In the above example, rearrangement gives
   f(x) =
   
   x − 3
   2
   2
   − 9
   4

• 2 =
  
  x − 3
  2
  2
  − 1
  4,
  from which we deduce that the axis of symmetry is x−3/2 = 0 and the vertex
  is (3/2,−1/4). Next consider the function f(x) = 2x−x2. This expression can
  be rearranged as follows to determine the axis of symmetry and vertex:
  f(x) = 2x − x2
  = −(x − 1)2 + 1
  Therefore, for this function we have a = −1, h = 1, and k = 1, so the axis
  of symmetry is the line x = 1 and the vertex is located at the point with
  coordinates (1, 1).
  Finally, we consider the function f(x) = 2×2 + 3x − 2. As before we write
  f(x) = 2×2 + 3x − 2
  = 2

x2 +
3
2x − 1

= 2

x +
3
4
2
− 9
16
− 1

= 2

x +
3
4
2
− 25
8
Therefore, for this function we have a = 2, h = −3/4, and k = −25/8, so the
axis of symmetry is the line x = −3/4 and the vertex is located at the point
with coordinates (−3/4,−25/8). In Table 3.5, we provide the axes and vertices
of the four quadratic functions we have investigated in this chapter.
56 Elements of Mathematics for Economics and Finance
Table 3.5 Axes and vertices of some quadratic functions.
f(x) Axis Vertex
x2 x = 0 (0, 0)
2x − x2 x = 1 (1, 1)
2×2 + 3x − 2 x = −3/4 (−3/4,−25/8)
x2 − 2x + 2 x = 1 (1, 1)
3.3 Quadratic Equations
There are a number of techniques for determining the roots of a quadratic
equation. Knowledge of the roots of a quadratic equation can be an additional
aid to sketching the graph of a quadratic function. If the expression defining
a quadratic function can be factorised as a product of linear factors, then
equating each of the factors to zero and solving the resulting linear equations
will provide the roots.
Example 3.1
Solve x2 + 13x + 30 = 0 using factorization.
Solution. First, we factorize the quadratic expression x2 + 13x + 30 as a
product of two linear factors (x + A) and (x + B), where A and B are two
constants that need to be determined. Since
(x + A)(x + B) = x2 + (A + B)x + AB,
then the constants A and B need to be chosen so that
A + B = 13, AB= 30.
The possible combinations of integers whose product is 30 are 30 × 1, 15 × 2,
10×3, and 6×5. Of course, one also has the combinations in which the integers
have been negated such as (−30)×(−1), but out of these combinations the only
one for which the pair of integers sums to 13 is 10 × 3. Therefore, we choose
A = 10 and B = 3, i.e.,
x2 + 13x + 10 = (x + 10)(x + 3).
We now solve the equation (x+3)(x + 10) = 0. For the product of the two
linear terms x+3 and x+10 to be zero, at least one of them must be zero. So
either x + 3 = 0 or x + 10 = 0.

3. Quadratic Equations 57
   If x + 3 = 0 then x = −3, and if x + 10 = 0 then x = −10. Therefore, the
   roots of the equation x2 + 13x + 30 = 0 are x = −3 and x = −10.
   Example 3.2
   Solve the quadratic equation 2×2 − 11x + 12 = 0 using factorization.
   Solution. As in the previous example, the first step is to factorize the
   quadratic expression 2×2 − 11x + 12 as a product of linear factors. These linear
   factors must be of the form (2x + A) and (x + B) in order to retrieve the
   quadratic factor 2×2, where A and B are two positive constants. Since
   (2x + A)(x + B) = 2×2 + (A + 2B)x + AB,
   then the constants A and B need to be chosen so that
   A + 2B = −11, AB= 12.
   The possible combinations of integers whose product is 12 are 12 × 1, 6 × 2,
   4 × 3, −4× −3, −6× −2, and −12 × −1. The only pair of integers amongst
   these for which A + 2B = −11 is A = −3 and B = −4. Therefore, we have
   2×2 − 11x + 12 = (2x − 3)(x − 4).
   The problem now is to solve the equation
   (2x − 3)(x − 4) = 0.
   Either 2x − 3 = 0 or x − 4 = 0. If 2x − 3 = 0 then 2x = 3 and x = 3/2. If
   x−4 = 0, then x = 4. Therefore, the two roots of the equation 2×2−11x+12 = 0
   are x = 3/2 and x = 4.
   Most quadratic expressions, however, do not factorise easily in the sense that
   they cannot be expressed as a product of linear factors with integer coefficients,
   even if the coefficients of the quadratic equation are integers. For example, the
   quadratic equation 3×2 − 9x + 5 = 0 cannot be factored into a product of
   linear factors with integer coefficients. Clearly, a more systematic approach is
   required.
   There is a formula for finding the solution to a quadratic equation
   ax2 + bx + c = 0. (3.7)
   58 Elements of Mathematics for Economics and Finance
   The formula may be derived by the process known as completing the square
   that was introduced in Section 3.2. We assume that a = 0. Using (3.6) we see
   that (3.7) is equivalent to
   a
   
   x + b
   2a
   2
   +
   4ac − b2
   4a
   = 0.
   Dividing both sides by a and taking the last term to the right-hand side yields
   
   x + b
   2a
   2
   = b2 − 4ac
   4a2 .
   Now taking the square root of both sides gives
   x + b
   2a
   = ±
   
   b2 − 4ac
   4a2 = ±
   √
   b2 − 4ac
   2a
   .
   Finally, subtracting b/(2a) from both sides we arrive at the formula for the
   roots of a quadratic equation:
   x =
   −b ±
   √
   b2 − 4ac
   2a
   . (3.8)
   This is an important formula for the roots (that is, solutions) of a quadratic
   equation, which we highlight:
   The solutions of the quadratic equation ax2 + bx + c = 0 are
   x =
   −b ±
   √
   b2 − 4ac
   2a
   .
   The number of solutions of a quadratic equation depends on the sign of the
   expression under the square root sign in this formula. A quadratic equation
   has two, one or no solutions depending on whether the expression b2 − 4ac is
   positive, zero, or negative:
   • If b2 − 4ac > 0, there are two solutions
   x =
   −b +
   √
   b2 − 4ac
   2a
   and x =
   −b −
   √
   b2 − 4ac
   2a
   .
   • If b2 − 4ac = 0, then there is one solution
   x = − b
   2a
   .
   • If b2 − 4ac < 0, then there are no solutions since the square root of b2 − 4ac
   does not exist in this case.
4. Quadratic Equations 59
   Example 3.3
   Solve the quadratic equation
   4×2 − 11x + 6 = 0
   using the formula.
   Solution. Compare the coefficients of this equation with those of the general
   quadratic equation. If we do this, we notice that a = 4, b = −11, and c = 6.
   Inserting these values into the formula (3.8) gives
   x =
   −(−11) ±
   
   (−11)2 − 4 × 4 × 6

2 × 4

11 ±
√
121 − 96

8

11 ±
√
25

8

11 ± 5
8
Therefore, the two solutions are
x =
11 + 5

8

16
8
= 2, and x =
11 − 5

8

6

8

3
4.
Example 3.4
Solve the quadratic equation
x2 − 2x − 15 = 0
using the formula.
Solution. Compare the coefficients of this equation with those of the general
quadratic equation. If we do this, we notice that a = 1, b = −2, and c = −15.
Inserting these values into the formula (3.8) gives
x =
−(−2) ±

(−2)2 − 4 × 1 × (−15)

2 × 1

2 ±

4 − (−60)

2

2 ±
√
64

2

2 ± 8
2
60 Elements of Mathematics for Economics and Finance
Therefore, the two solutions are
x =
2 + 8

2

10
2
= 5, and x =
2 − 8

2

−6
2
= −3.
Example 3.5
Solve the quadratic equation
3×2 − 9x + 5 = 0
using the formula.
Solution. Compare the coefficients of this equation with those of the general
quadratic equation. If we do this, we notice that a = 3, b = −9, and c = 5.
Inserting these values into the formula (3.8) gives
x =
−(−9) ±

(−9)2 − 4 × 3 × 5

2 × 3

9 ±
√
81 − 60

6

9 ±
√
21
6
Note that 21 is not a perfect square, and therefore the roots of this equation can
only be expressed in decimal representation to a specified number of decimal
places. Therefore, to four decimal places the two solutions of this equation are
x =
9 +
√
21
6
= 2.2638, and x =
9 −
√
21
6
= 0.7362.
Example 3.6
Solve the quadratic equation x2 − 18x + 45 = 0 by completing the square.
Solution. In this example a = 1, b = −18, and c = 45. Therefore, using (3.6)
we may write the equation in the form
(x − 9)2 − 81 + 45 = 0,
or
(x − 9)2 = 36.
Then taking the square root of both sides gives
x −9 = ±6.

3. Quadratic Equations 61
   Either x − 9 = 6, which means that x = 15. Or x −9 = −6, which means
   that x = 3.
   3.4 Applications to Economics
   One function of particular interest in economics is the profit function. We
   denote this function by the Greek symbol π. The profit function is defined to
   be the difference between total revenue, TR, and the total cost, TC, i.e.,
   π = TR − TC.
   The total revenue received from the sale of Q goods at price P is given by the
   product of P and Q, i.e.,
   TR = P × Q.
   The total cost function relates the cost of production to the level of output, Q,
   and is the sum of the fixed costs, FC, and variable costs, V C × Q, where V C
   denotes the variable cost per unit of output. Fixed costs include, for example,
   the cost of land, rental, equipment, and skilled labour. Variable costs include,
   for example, the cost of raw materials, energy, and unskilled labour. The total
   cost in producing Q goods is given by
   TC = FC + (V C) × Q.
   Thus the profit function is
   π = P × Q − [FC + (V C) × Q] = PQ − FC − (V C) × Q.
   Note that care needs to be exercised in removing the brackets. It is important
   to remember that the negative sign outside the square brackets negates all
   terms inside the brackets when the brackets are moved.
   Example 3.7
   If fixed costs are 18, variable costs per unit are 4, and the demand function is
   P = 24 − 2Q
   obtain an expression for π in terms of Q and hence sketch a graph of π against
   Q.
4. For what values of Q does the firm break even?
5. What is the maximum profit?
   62 Elements of Mathematics for Economics and Finance
   Table 3.6 Table of values of the profit function π = −2Q2 +20Q−18 for
   even integer values of Q for which 0 ≤ Q ≤ 10.
   Q 0 2 4 6 8 10
   −2Q2 0 −8 −32 −72 −128 −200
   20Q 0 40 80 120 160 200
   −18 −18 −18 −18 −18 −18 −18
   π −18 14 30 30 14 −18
   Solution. The total revenue function is given by
   TR = P × Q = (24 − 2Q)Q = 24Q − 2Q2,
   where we have used the demand function P = 24 − 2Q to eliminate P in the
   expression defining TR. We have expressed TR solely in terms of the level of
   output, Q. The total cost function is given by
   TC = FC + (V C) × Q = 18+4Q,
   since FC = 18 and V C = 4. We can now obtain an expression for the profit
   function by subtracting the expression for TC from the expression for TR, i.e.,
   π = TR − TC
   = 24Q − 2Q2 − (18 + 4Q)
   = 24Q − 2Q2 − 18 − 4Q
   = −2Q2 + 20Q − 18,
   where we have taken care to change the sign of all terms inside the brackets on
   their removal.
   Since the coefficient of Q2 in the quadratic expression defining π is negative,
   the graph of the profit function has a
   
   shape. When Q = 0, π = −18. The
   profit function is tabulated in Table 3.6 for 0 ≤ Q ≤ 10. From this information,
   we are able to sketch the graph of the function. This is shown in Fig. 3.5.
6. The value of the profit function will be zero (i.e., π = 0) for values of Q
   that satisfy the quadratic equation
   −2Q2 + 20Q − 18 = 0.
7. Quadratic Equations 63
   Solving this equation using the formula with a = −2, b = 20, and c = −18
   yields
   Q =
   −20 ±
   √
   400 − 144

−4

−20 ±
√
256

−4

−20 ± 16
−4
Therefore, either
Q =
−20 + 16

−4

−4
−4
= 1,
or
Q =
−20 − 16

−4

−36
−4
= 9.
The profit is zero when Q = 1 and Q = 9. Therefore, the firm breaks even
when Q = 1 and Q = 9. For 1 < Q < 9, the profit function is positive (see Fig. 3.5) and the firm is in profit. For values of Q outside this range, i.e., Q < 1 and Q > 9, the profit function is negative and therefore the firm
makes a loss at these levels of output.
Q
0 2 4 6 8 10 12
-50
-30
-10
10
30
π (5,32)
Figure 3.5 The graph of the profit function π = −2Q2 + 20Q − 18.
64 Elements of Mathematics for Economics and Finance

2. To determine the maximum value of the profit function, we complete the
   square.
   π = −2

Q2 − 10Q + 9

= −2

(Q − 5)2 − 25 + 9

= −2

(Q − 5)2 − 16

= (−2) × (Q − 5)2 + (−2) × (−16)
= −2(Q − 5)2 + 32
Therefore, the maximum profit is π = 32 since the term −2(Q − 5)2 is
always negative except when Q = 5 when it is zero.
Finally, we return to supply and demand analysis. In Chapter 2, we considered
examples in which both the supply and demand functions were linear
and determined the equilibrium price and quantity. Although linear models are
frequently used in economics because of the simplicity of their mathematical
structure, they can also be limiting in the sort of economic behaviour they
describe. As we shall see in the next example, it is not necessary for the supply
and demand functions to be linear, and, in the case when they are defined by
quadratic expressions, the market equilibrium can be determined by solving a
quadratic expression.
Example 3.8
Given the supply and demand functions
P = Q2s+ 12Qs + 32,
P = −Q2
d
− 4Qd + 200,
calculate the equilibrium price and quantity.
Solution. At equilibrium, the quantity supplied is equal to the quantity demanded,
so that
Qd = Qs = Q, say.
Then the supply and demand equations become
P = Q2 + 12Q + 32,
P = −Q2 − 4Q + 200.
Equating the expressions on the right-hand sides of these equations, we have
Q2 + 12Q + 32 = −Q2 − 4Q + 200.

3. Quadratic Equations 65
   Q
   -15 -10 -5 0 5 10
   50
   100
   150
   200
   250
   Supply Function
   Demand Function
   P
   economically irrelevant economically relevant
   region region
   Figure 3.6 The graph of the supply and demand functions in Example
   3.8.
   We can do this since both expressions are equal to P. Rearranging this equation
   and collecting like terms yields the quadratic equation
   2Q2 + 16Q − 168 = 0.
   This equation can be simplified by dividing throughout by 2. We then have the
   quadratic equation
   Q2 + 8Q − 84 = 0.
   Solving this equation using the formula with a = 1, b = 8, and c = −84 yields
   Q =
   −8 ±
   
   82 − 4 × 1 × (−84)

2 × 1

−8 ±
√
64 + 336

2

−8 ±
√
400

2

−8 ± 20
2
Therefore, either
Q =
−8 + 20

2

12
2
= 6,
66 Elements of Mathematics for Economics and Finance
or
Q =
−8 − 20

2

−28
2
= −14.
So the quadratic equation has solutions Q = 6 and Q = −14. The solution
Q = −14 can be discarded because a negative quantity does not make sense.
Therefore, the equilibrium quantity is 6. The corresponding equilibrium price
can be determined by substituting Q = 6 into either the supply or demand
equation. If we substitute this value into the supply equation, we have
P = 62 + 12 × 6 + 32 = 36 + 72 + 32 = 140.
Therefore, the equilibrium price is 140.
The graphs of the supply and demand functions are shown in Fig. 3.6.
There are two points of intersection. The one for positive Q provides the market
equilibrium.
EXERCISES
3.1. Evaluate the function f(x) = 2×2−9x+4 when x = 0, 1, 2, 3, 4, 5.
Hence, sketch the graph of this function for 0 ≤ x ≤ 5.
3.2. Evaluate the function f(x) = −2×2 − 3x + 3 when x = −3, −2, −1,
0, 1, 2. Hence, sketch the graph of this function for −3 ≤ x ≤ 2.
3.3. Sketch the graphs of the following functions:
a) f(x) = 4×2 − 7x − 2, for −2 ≤ x ≤ 4; ,
b) f(x) = 9 − 6x − 8×2, for −3 ≤ x ≤ 3.
3.4. Solve the following quadratic equations using the formula:
a) x2 − 4x + 3 = 0,
b) 3×2 + 5x − 8 = 0,
c) 2×2 − 19x − 10 = 0.
3.5. Solve the following quadratic equations using factorization:
a) x2 + 7x + 10 = 0,
b) x2 − 4x − 5 = 0,
c) 6×2 + 19x + 10 = 0.
3.6. Write the quadratic function f(x) = x2 − 8x + 12 in the form
f(x) = a(x − h)2 + k.

3. Quadratic Equations 67
   What is the equation for the axis of symmetry of this parabola, and
   what is its vertex? Use this information to sketch the graph of this
   function.
   3.7. If fixed costs are 6, variable costs per unit are 2, and the demand
   function is
   P = 15 − 3Q
   obtain an expression for the profit function π in terms of Q. Hence,
   sketch a graph of π against Q.
   3.8. If fixed costs are 4, variable costs per unit are 3, and the demand
   function is
   P = 45 − 4Q
   obtain an expression for the profit function π in terms of Q. Hence,
   sketch a graph of π against Q.