IB Mathematics SL 2.5 Linear models AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Ava hits a baseball. The height of the baseball after it is hit by a bat is modelled by the function:
\[ h(t) = -4.8 t^2 + 21 t + 1.2 \]
where \( h(t) \) is the height in metres above the ground and \( t \) is the time in seconds after the ball was hit.
(a) Write down the height of the ball above the ground at the instant it is hit by the bat. [1]
(b) Find the value of \( t \) when the ball hits the ground. [2]
(c) State an appropriate domain for \( t \) in this model. [1]
▶️ Answer/Explanation
Markscheme
(a)
At \( t = 0 \):
\[ h(0) = -4.8(0)^2 + 21(0) + 1.2 = 1.2 \text{ m} \] A1
Height: \( 1.2 \text{ m} \).
[1 mark]
At \( t = 0 \):
\[ h(0) = -4.8(0)^2 + 21(0) + 1.2 = 1.2 \text{ m} \] A1
Height: \( 1.2 \text{ m} \).
[1 mark]
(b)
Set \( h(t) = 0 \):
\[ -4.8 t^2 + 21 t + 1.2 = 0 \] M1
Solve using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = -4.8 \), \( b = 21 \), \( c = 1.2 \):
\[ \begin{aligned} \text{Discriminant} &= 21^2 – 4(-4.8)(1.2) \\ &= 441 + 23.04 = 464.04 \end{aligned} \]
\[ \begin{aligned} t &= \frac{-21 \pm \sqrt{464.04}}{2(-4.8)} \\ &= \frac{-21 \pm \sqrt{464.04}}{-9.6} \\ t &\approx \frac{-21 + 21.5414}{-9.6} \approx 4.431415 \text{ s} \end{aligned} \] A1
Time: \( t \approx 4.43 \text{ s} \).
[2 marks]
Set \( h(t) = 0 \):
\[ -4.8 t^2 + 21 t + 1.2 = 0 \] M1
Solve using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = -4.8 \), \( b = 21 \), \( c = 1.2 \):
\[ \begin{aligned} \text{Discriminant} &= 21^2 – 4(-4.8)(1.2) \\ &= 441 + 23.04 = 464.04 \end{aligned} \]
\[ \begin{aligned} t &= \frac{-21 \pm \sqrt{464.04}}{2(-4.8)} \\ &= \frac{-21 \pm \sqrt{464.04}}{-9.6} \\ t &\approx \frac{-21 + 21.5414}{-9.6} \approx 4.431415 \text{ s} \end{aligned} \] A1
Time: \( t \approx 4.43 \text{ s} \).
[2 marks]
(c)
The ball is hit at \( t = 0 \) and lands at \( t \approx 4.43 \text{ s} \).
Domain: \( [0, 4.43] \). A1
[1 mark]
The ball is hit at \( t = 0 \) and lands at \( t \approx 4.43 \text{ s} \).
Domain: \( [0, 4.43] \). A1
[1 mark]
Total Marks: 4
Question
Consider the function f(x) = ax² + bx + c. The graph of y = f(x) is shown in the diagram. The vertex of the graph has coordinates (0.5, -12.5). The graph intersects the x-axis at two points, (-2, 0) and (p, 0).
The graph of y = f(x) is a parabola with vertex (0.5, -12.5) and x-intercepts (-2, 0) and (p, 0).
(a) Determine the value of p. [1]
(b) Determine the value of
(i) a.
(ii) b.
(iii) c. [4]
(i) a.
(ii) b.
(iii) c. [4]
(c) Write down the equation of the axis of symmetry of the graph. [2]
▶️ Answer/Explanation
Markscheme
(a)
The vertex of the parabola lies on the axis of symmetry, which is the midpoint of the x-intercepts at \( x = -2 \) and \( x = p \).
Given the vertex x-coordinate is 0.5, calculate: \( \frac{-2 + p}{2} = 0.5 \).
Solve: \( -2 + p = 1 \), so \( p = 1 + 2 = 3 \).
Thus, the value of p is 3. A1
[1 mark]
The vertex of the parabola lies on the axis of symmetry, which is the midpoint of the x-intercepts at \( x = -2 \) and \( x = p \).
Given the vertex x-coordinate is 0.5, calculate: \( \frac{-2 + p}{2} = 0.5 \).
Solve: \( -2 + p = 1 \), so \( p = 1 + 2 = 3 \).
Thus, the value of p is 3. A1
[1 mark]
(b)
Use the quadratic function f(x) = ax² + bx + c with vertex (0.5, -12.5) and x-intercepts (-2, 0) and (3, 0).
Method 1 (System of Equations):
Set up equations using the given points:
(1) At \( x = -2 \), \( f(-2) = 0 \): \( 4a – 2b + c = 0 \).
(2) At \( x = 3 \), \( f(3) = 0 \): \( 9a + 3b + c = 0 \).
(3) At vertex \( x = 0.5 \), \( f(0.5) = -12.5 \): \( \frac{1}{4}a + \frac{1}{2}b + c = -12.5 \).
Subtract (1) from (2): \( (9a + 3b + c) – (4a – 2b + c) = 0 \), so \( 5a + 5b = 0 \), \( a + b = 0 \), \( b = -a \). M1
Substitute \( b = -a \) into (1): \( 4a – 2(-a) + c = 4a + 2a + c = 6a + c = 0 \).
Substitute \( b = -a \) into (3): \( \frac{1}{4}a + \frac{1}{2}(-a) + c = \frac{1}{4}a – \frac{1}{2}a + c = -\frac{1}{4}a + c = -12.5 \).
Solve the system: \( 6a + c = 0 \), \( -\frac{1}{4}a + c = -12.5 \).
Subtract: \( (6a + c) – \left(-\frac{1}{4}a + c\right) = 0 – (-12.5) \), \( 6a + \frac{1}{4}a = 12.5 \), \( \frac{24a + a}{4} = \frac{25a}{4} = 12.5 \), \( 25a = 50 \), \( a = 2 \).
Then: \( 6 \times 2 + c = 12 + c = 0 \), \( c = -12 \).
And: \( b = -a = -2 \).
Method 2 (Factor Form):
Use the factor form: f(x) = a(x + 2)(x – 3).
Substitute vertex (0.5, -12.5): \( f(0.5) = a(0.5 + 2)(0.5 – 3) = a \times 2.5 \times (-2.5) = -6.25a = -12.5 \).
Solve: \( -6.25a = -12.5 \), \( a = 2 \).
Expand: f(x) = 2(x + 2)(x – 3) = 2(x² – x – 6) = 2x² – 2x – 12, so a = 2, b = -2, c = -12.
Thus: (i) a = 2, (ii) b = -2, (iii) c = -12. A1 A1 A1
[4 marks]
Use the quadratic function f(x) = ax² + bx + c with vertex (0.5, -12.5) and x-intercepts (-2, 0) and (3, 0).
Method 1 (System of Equations):
Set up equations using the given points:
(1) At \( x = -2 \), \( f(-2) = 0 \): \( 4a – 2b + c = 0 \).
(2) At \( x = 3 \), \( f(3) = 0 \): \( 9a + 3b + c = 0 \).
(3) At vertex \( x = 0.5 \), \( f(0.5) = -12.5 \): \( \frac{1}{4}a + \frac{1}{2}b + c = -12.5 \).
Subtract (1) from (2): \( (9a + 3b + c) – (4a – 2b + c) = 0 \), so \( 5a + 5b = 0 \), \( a + b = 0 \), \( b = -a \). M1
Substitute \( b = -a \) into (1): \( 4a – 2(-a) + c = 4a + 2a + c = 6a + c = 0 \).
Substitute \( b = -a \) into (3): \( \frac{1}{4}a + \frac{1}{2}(-a) + c = \frac{1}{4}a – \frac{1}{2}a + c = -\frac{1}{4}a + c = -12.5 \).
Solve the system: \( 6a + c = 0 \), \( -\frac{1}{4}a + c = -12.5 \).
Subtract: \( (6a + c) – \left(-\frac{1}{4}a + c\right) = 0 – (-12.5) \), \( 6a + \frac{1}{4}a = 12.5 \), \( \frac{24a + a}{4} = \frac{25a}{4} = 12.5 \), \( 25a = 50 \), \( a = 2 \).
Then: \( 6 \times 2 + c = 12 + c = 0 \), \( c = -12 \).
And: \( b = -a = -2 \).
Method 2 (Factor Form):
Use the factor form: f(x) = a(x + 2)(x – 3).
Substitute vertex (0.5, -12.5): \( f(0.5) = a(0.5 + 2)(0.5 – 3) = a \times 2.5 \times (-2.5) = -6.25a = -12.5 \).
Solve: \( -6.25a = -12.5 \), \( a = 2 \).
Expand: f(x) = 2(x + 2)(x – 3) = 2(x² – x – 6) = 2x² – 2x – 12, so a = 2, b = -2, c = -12.
Thus: (i) a = 2, (ii) b = -2, (iii) c = -12. A1 A1 A1
[4 marks]
(c)
The axis of symmetry of a quadratic f(x) = ax² + bx + c is given by \( x = -\frac{b}{2a} \).
From (b), a = 2, b = -2, so: \( x = -\frac{-2}{2 \times 2} = \frac{2}{4} = 0.5 \).
Alternatively, the vertex at (0.5, -12.5) directly gives the axis of symmetry as \( x = 0.5 \). A1 A1
[2 marks]
The axis of symmetry of a quadratic f(x) = ax² + bx + c is given by \( x = -\frac{b}{2a} \).
From (b), a = 2, b = -2, so: \( x = -\frac{-2}{2 \times 2} = \frac{2}{4} = 0.5 \).
Alternatively, the vertex at (0.5, -12.5) directly gives the axis of symmetry as \( x = 0.5 \). A1 A1
[2 marks]
Total Marks: 7