Consider the general quadratic equation:
y = ax2 + bx + c
By completing the square, this equation may be manipulated into the form y = a(x − b)2 + c
where the turning point is (b, c).
This way of writing the function is known as the power form or turning point form. The transformations associated with this form will be discussed more fully in chapter 2.
Worked exAMple 19
For the function y==−−2(x++3)2−−4, find:
a the coordinates of the turning point b the domain and range.
ThINk WrITe
Write the general formula. y = a(x − b)2 + c
Write the function. y = −2(x + 3)2 − 4
a 1 Identify the values of a, b and c. a a = −2, b = −3, c = −4
2 State the coordinates of the turning point (b, c).
The turning point is (−3, −4).
b 1 Write the domain of the parabola. b The domain is R.
2 Write the range y ≤ c (as a < 0). The range is y ≤ −4.
Worked exAMple 20
The function graphed at right is of the form y x== 2++bx c++ . Find:
a the rule b the domain c the range.
Write the answers to b and c in interval notation.
ThINk WrITe
a 1 Write the general rule for a quadratic in turning point form.
a y = a(x − b)2+ c
2 Find the values of b and c using the given turning point.
Since the turning point is (−1, −6):
b = −1, c = −6
3 State the value of a (given). a = 1
4 Substitute these values in the rule. So y = 1(x + 1)2 − 6
5 Expand the brackets. = x2+ 2x + 1 − 6
6 Simplify. = x2+ 2x − 5
The rule is y = x2+ 2x − 5.
b 1 use the graph to find the domain.
Look at all the values that x can take.
b x ≥ −5
2 State the domain in interval notation. Domain = [−5, ∞) c 1 use the graph to find the range. Look
at all the values that y can take.
c y ≥−6
2 State the range in interval notation. Range = [−6, ∞)
Worked exAMple 21
Sketch the graph of y==12(x−−1)2++2, clearly showing the coordinates of the turning point and the intercepts with the axes. State its range.
ThINk WrITe/drAW
1 Write the general equation of the parabola. y = a(x − b)2 + c
2 Identify the values of the variables. a = 12, b = 1, c = 2 y
0 x
(−1, −6) (−5, 10)
3 Write a brief statement on the transformation of the basic parabola.
The graph of y = x2 is dilated in the y direction by the factor of 12 (that is, it is wider than the basic curve); it is translated 1 unit to the right and 2 units up.
4 State the shape of the parabola (that is, positive or negative).
a > 0; the parabola is positive.
5 State the coordinates of the turning point (b, c). The turning point is (1, 2).
6 As both a and c are positive, only the y-intercept needs to be determined. Find the y-intercept by making x = 0.
y-intercept: x = 0 y = 12(0 − 1)2 + 2
= 12(−1)2+ 2
= 12 + 2
= 212
7 Sketch the graph:
Draw a set of axes and label them. Plot the turning point and the y-intercept. Sketch the graph of the positive parabola, so that it passes through the points previously marked.
y
0 1 x 21–2
y = (x − 1)1–2 2+ 2 2
8 Since y ≥ 2, that is the range. The range is y ∈ [2, ∞).
Worked exAMple 22
Sketch the graph of y== ++3 8x−−2x2, showing the turning point and all intercepts, rounding answers to 2 decimal places where appropriate.
ThINk WrITe/drAW
Method 1: Technology-free
1 Find y when x = 0. When x = 0, y = 3
2 State the y-intercept. The y-intercept is 3.
3 Let the quadratic equal zero. When y = 0,
3 + 8x − 2x2 = 0
4 Solve for x using the quadratic formula.
x= − ± − −
−
8 8 4 2 3
2 2
2 ( )( ) ( )
= − ±
−
8 88
4
= − ±
−
8 2 22 4
= − ±
−
4 22
2
= −2 22
2 or 2 22
+ 2
5 State the x-intercepts, rounding to 2 decimal places.
The x-intercepts are −0.35 and 4.35.
6 use the formula for the x-value of the turning point, x b
= −2a. x= −−8
2 2( ) x = 2
7 To calculate the y-coordinate of the turning point,
substitute x = 2 into the function. y = −2(2)2+ 8(2) + 3 y = 11
8 State the turning point. The turning point is (2, 11).
9 Draw a set of axes and mark the coordinates of the turning point and the points where the graph crosses the axes.
y
0 4 x
–1 5
(2, 11)
(–0.35, 0) f(x) = 3 + 8x – 2x2
(4.35, 0) 3
6 9 12
(0, 3) 10 Sketch a parabola through these points.
Method 2: Technology-enabled
Note: Function notation includes the rule, the domain and the co-domain. For example, f (x): [−2, 1] → R, where f (x) = x2 − 3, is a parabola with rule f (x) = x2 − 3 and domain [−2, 1].
The range is a subset of the co-domain, R.
Worked exAMple 23
The weight of a person t months after a gymnasium program is started is given by the function: W t( ) t t
22 3 80
== −− + , where t ∈ [0, 8]
and W is in kilograms. Find:
a the minimum weight of the person b the maximum weight of the person.
On a Graphs page, complete the function entry line as:
f(x) = 3 + 8x − 2x2 Then press ENTER ·.
To label the coordinates of the intercepts and turning point, press:
• MENu b
• 6:Analyze Graph 6 Select the appropriate action.
ThINk WrITe 1 Complete the square to find the turning point. W t
t
t t
t t
= − +
= − +
= − + + −
2
1 2 2 1 2 2
2 3 80
6 160
6 9 160 9
[ ]
[ ]
== − +
= − +
1
2 2
1
2 2
3 151
3 75 5
[( ) ]
( ) .
t t
2 State the minimum turning point. The turning point is (3, 75.5).
3 Find the end point value for W when t = 0. When t = 0, W = 80
4 State its coordinates. One end point is (0, 80).
5 Find the end point value of W when t = 8. When t = 8, W = 88
6 State its coordinates. The other end point is (8, 88).
7 On a set of axes, mark the end points and turning point. W (kg)
t (months) 0 1 2 3 4 5 6 7 8
90 80 70
Maximum (8, 88)
Minimum (3, 75.5) (0, 80)
8 Sketch a parabola between the end points.
9 Locate the maximum and minimum values of W on the graph.
a State the minimum weight from the graph. a The minimum weight is 75.5 kg.
b State the maximum weight from the graph. b The maximum weight is 88 kg.
Quadratic graphs
Quadratic equations are polynomials of degree 2.
1.
The general equation is
2. y = ax2+ bx + c.
The quadratic formula is given by the equation 3.
x b b ac
= − ± a2 4− 2 The discriminant is
4. b2 − 4ac and if:
(a) b2 − 4ac > 0, there are two x-intercepts. If b2 − 4ac is a perfect square, the intercepts are rational.
(b) b2 − 4ac = 0, there is one x-intercept, which is a turning point.
(c) b2 − 4ac < 0, there are no x-intercepts.
The turning point form of the quadratic graph or parabola is:
5.
y = a(x − b)2+ c and the turning point is (b, c).
The equation of the axis of symmetry of a parabola and the
6. x-value of the turning
point is given by the expression x = −b a 2 . The axis of symmetry is halfway between the
7. x-intercepts.
reMeMBer
Quadratic graphs
1 We17 use the discriminant to determine the number of x-intercepts for each of the following quadratic functions.
a f (x) = x2 − 3x + 4 b f (x) = x2 + 5x − 8 c f (x) = 3x2 − 5x + 9 d f (x) = 2x2 + 7x − 11 e f (x) = 1 − 6x − x2 f f (x) = 3 + 6x + 3x2 2 We18 Sketch the graphs of each of the following functions,
showing all intercepts. Give exact answers.
a f (x) = x2 − 6x + 8 b f (x) = x2 − 5x + 4 c f (x) = 10 + 3x − x2 d f (x) = 6x2 − x − 12
3 Find the turning point for each of the functions in question 2.
Give exact answers.
4 We19 For each of the following functions find:
i the coordinates of the turning point ii the domain iii the range.
a y = 2 − x2 b y = (x − 6)2 c y = −(x + 2)2 d y = 2(x + 3)2 − 6
5 We20 Each of the functions graphed below is of the form y = x2 + bx + c. For each function, give: i the rule ii the domain iii the range.
Write the answers to b and c in interval notation.
a y
0 x (1,−2)
b y
0 x (−1, 6)
(2, −3)
c y
0 x (1, 9)
(−4, −16) 4
6 We21 Sketch the graphs of the following, clearly showing the coordinates of the turning point and the intercepts with the axes.
a y = 2x2+ 3 b y = (2x − 5)(2x − 3) c y = (2x − 3)2 − 8 7 MC Consider the function with the rule y = x2 − 2x − 3.
a It has x-intercepts:
A (1, 0) and (3, 0) B (−1, 0) and (3, 0) C (1, 0) and (−3, 0) D (2, 0) and (−1, 0) E (0, −1) and (0, 3)
b It has a turning point with coordinates:
A (−1, 0) B (2, −3) C (1, −4) D (−1, −4) E (1, 0) 8 MC The function f (x) = −(x + 3)2+ 4 has a range given by:
A (3, ∞) B (−∞, −3] C [4, ∞) D (−∞, 4] E R− 9 MC The range of the function y = (x − 4)2, x ∈ [0, 6] is:
A [0, 16] B [4, 16] C [0, 4] D (4, 12] E [0, 16) 10 We22 Sketch the graph of each of the following functions, showing
the turning point and all intercepts. Round answers to 2 decimal places where appropriate.
a f (x) = (x − 2)2 − 4 b f (x) = −(x + 4)2+ 9 c y = x2+ 4x + 3 d y = 2x2 − 4x − 6 exerCISe
1e
eBookplus eBookplus Digital doc Spreadsheet 103 Discriminant
eBookplus eBookplus Digital doc Spreadsheet 107 Quadratic graphs
eBookplus eBookplus Digital doc SkillSHEET 1.7 Domain and range for quadratic graphs
eBookplus eBookplus Digital doc Spreadsheet 041 Function grapher
eBookplus eBookplus Digital doc Spreadsheet 108 Quadratic graphs —
turning point form
11 Sketch the graph of each of the functions below and state i the domain and ii the range of each function.
a y = x2 − 2x + 2, x ∈ [−2, 2] b y = −x2 + x − 1, x ∈ R+ c f (x) = x2 − 3x − 2, x ∈ [−10, 6] d f (x) = 5 + 6x − 3x2, x ∈ [−5, 3)
12 We23 The volume of water in a tank, V m3, over a 10 month period is given by the function V(t ) = 2t 2 − 16 t + 40, where t is in months and t ∈ [0, 10].
Find:
a the minimum volume of water in the tank b the maximum volume of water in the tank.
13 A ball thrown upwards from a tower attains a height above the ground given by the function h(t) = 12t − 3t 2+ 36, where t is the time in seconds and h is in metres.
Find:
a the maximum height above the ground that the ball reaches
b the time taken for the ball to reach the ground
c the domain and range of the function.
14 A section of a roller-coaster at an amusement park follows the path of a parabola. The function h(t) = t2 − 12t + 48, t ∈ [0, 11], models the height above the ground of the front of one of the carriages, where t is the time in seconds and h is the height in metres.
a Find the lowest point of this section of the ride.
b Find the time taken for the carriage to reach the lowest point.
c Find the highest point above the ground.
d Find the domain and the range of the function.
e Sketch the function.