Quadratics IB Questions 4.
The diagram below shows the graph of y = c + kx – x2, where k and c are constants.
y
Q
O
P(5, 0)
(a)
Find the values of k and c.
(b)
Find the coordinates of Q, the highest point on the graph.
x
(Total 8 marks)
12.
The diagram shows the graph of y = x2 – 2x – 8. The graph crosses the x-axis at the point A, and has a
vertex at B.
y
A
x
O
B
(a)
Factorize x2 – 2x – 8.
(b)
Write down the coordinates of each of these points
(i)
A;
(ii)
B.
(Total 4 marks)
17.
The graph of the function y = x2 – x – 2 is drawn below.
y
A
0
B
x
C
(a)
Write down the coordinates of the point C.
(b)
Calculate the coordinates of the points A and B.
(Total 8 marks)
19.
Consider the graphs of the following functions.
(i)
y = 7x + x2;
(ii)
y = (x – 2)(x + 3);
(iii)
y = 3x2 – 2x + 5;
(iv)
y = 5 – 3x – 2x2.
Which of these graphs
(a)
has a y-intercept below the x-axis?
(b)
passes through the origin?
(c)
does not cross the x-axis?
(d)
could be represented by the following diagram?
y
O
x
(Total 8 marks)
21.
Diagram 1 shows a part of the graph of y = x2.
y
x
0
Diagram 1
Diagrams 2, 3 and 4 show a part of the graph of y = x2 after it has been moved parallel to the
x-axis, or parallel to the y-axis, or parallel to one axis, then the other.
y
y
y
3
3
0
Diagram 2
x
0 2
Diagram 3
x
0 2
x
Diagram 4
Write down the equation of the graph shown in
(a)
Diagram 2;
(b)
Diagram 3;
(c)
Diagram 4.
(Total 4 marks)
2.
(a)
Factorize 3x2 + 13x −10.
(2)
(b)
Solve the equation 3x2 + 13x − 10 = 0.
(2)
Consider a function f (x) = 3x2 + 13x −10.
(c)
Find the equation of the axis of symmetry on the graph of this function.
(2)
(d)
Calculate the minimum value of this function.
(2)
(Total 8 marks)
8.
(a)
Find the solution of the equation x2 – 5x – 24 = 0.
(b)
The equation ax2 – 9x – 30 = 0 has solution x = 5 and x = –2. Find the value of a.
(Total 8 marks)
Quadratics IB Answers 4.
(a)
y = x(5 – x) or y = 5x – x2 or 25 = c + 5k
c = 0, k = 5
(M1)
(A1)(
(C3)
A1)
Note: Award (A1) if no method is indicated but c = 0 or k = 5 is given
alone.
(b)
Vertex at x =
−b −5
=
= 2.5
2a − 2
(M1)
(A1)
y = 5(2.5) – 2.52 = 6.25
(M1)
(A1)
Note: The substitutions must be attempted to receive the method marks.
Q(2.5, 6.25)
(A1)
(C5)
Notes: Coordinate pair is required for (A1) but Q is not essential. If no
working shown and answer not fully correct, award (G2) for each
correct value and (A1) for coordinate brackets. However, if values are
close but not exactly correct (eg (2.49, 6.25)) award only (G1) for each
less precise value. In this case AP might also apply if number of digits is
inappropriate.
If differentiation is used, award (M1) for correct process, (A1) for x =
2.5, (M1)(A1) or (G2) for 6.25 and (A1) for coordinate brackets.
12.
(a)
(x + 2)(x – 4)
(A1)
(b)
(i)
(–2, 0)
(A1)
(ii)
(1, –9)
(A1)(A1)
[4]
17.
(a)
(b)
Put x = 0 to find y = –2
Coordinates are (0, –2)
Note: Award (M1)(A0) for –2 if working is shown. If not, award
(M0)(A0).
(M1)
(A1)
(C2)
Factorise fully, y = (x – 2) (x + 1).
(A1)(
A1)
y = 0 when x = –1, 2.
(A1)(
A1)
Coordinates are A(–1, 0), B(2, 0).
(A1)(
(C6)
A1)
Note: Award (C2) for each correct x value if no method shown and full
coordinates not given. If the quadratic formula is used correctly award
(M1)(A1)(A1)(A1)(A1)(A1). If the formula is incorrect award only the
last (A1)(A1) as ft.
[8]
19.
(a)
(ii)
(A2)
(C2)
(b)
(i)
(A2)
(C2)
(c)
(iii)
(A2)
(C2)
(d)
(iv)
(A2)
(C2)
[8]
21.
(a)
y = x2 + 3
(A1)
(b)
y = (x – 2)2
(A1)
(c)
y = (x – 2)2 + 3
(A2)
4
[4]
2.
Unit penalty (UP) is applicable where indicated.
(a)
(3x – 2)(x + 5)
(b)
(3x – 2)(x + 5) = 0
(c)
x=
2
or x = –5
3
x=
− 13
(− 2.17)
6
(A1)(A1)
2
(A1)(ft)(A1)(ft)(G2)
2
(A1)(A1)(ft)(G2)
2
Note: (A1) is for x =, (A1) for value. (ft) if
value is half way between roots in (b).
2
(d)
⎛ − 13 ⎞
⎛ − 13 ⎞
Minimum y = 3 ⎜
⎟ + 13⎜
⎟ − 10
⎝ 6 ⎠
⎝ 6 ⎠
Note: (M1) for substituting their value of x from
(c) into f(x)
= −24.1
(M1)
(A1)(ft)(G2)
2
[8]
8.
(a)
(b)
(x – 8)(x + 3) = 0
x = 8, x = –3
(M1)(M1)
(A1)(A1)
(C2)(C2)
METHOD 1
(x – 5)(x + 2) = 0
x2 – 3x – 10 = 0
3x2– 9x – 30 = 0
a=3
(M1)
(A1)
(A1)
(A1)
(C4)
METHOD 2
a(5)2 – 9(5) – 30 = 0
25a – 75 = 0
a=3
(M1)
(A2)
(A1)
(C4)
METHOD 3
a(–2)2 – 9(–2) – 30 = 0
4a – 12 = 0
a=3
(M1)
(A2)
(A1)
(C4)
[8]