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IB Mathematics SL 2.6 The quadratic function AA SL Paper 1- Exam Style Questions- New Syllabus

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Question

The function \( f \) is defined by \( f(x) = 5(x + 1)(x + 3) \), for \( x \in \mathbb{R} \).
(a) Express \( f(x) \) in the form \( a(x – h)^2 + k \), where \( a, h, k \in \mathbb{Z} \).
(b) Sketch the graph of \( y = f(x) \), indicating the coordinates of the vertex and the values of any intercepts with the axes.
(c) Solve the inequality \( f(x) \leq 40 \).
The function \( g \) is defined by \( g(x) = \ln x \), for \( x \in \mathbb{R}, x > 0 \).
(d) (i) Write down an expression for the composite function \( (f \circ g)(x) \).
 (ii) Hence, solve the inequality \( (f \circ g)(x) \leq 40 \).

Most-appropriate topic codes (Mathematics: analysis and approaches guide):

• SL 2.6: The quadratic function \( f(x) = ax^2 + bx + c \)… the form \( f(x) = a(x – h)^2 + k \), vertex \( (h, k) \) — Part a, b
• SL 2.4: Determine key features of graphs… Maximum and minimum values; intercepts; symmetry; vertex; zeros — Part b
• SL 2.7: Solution of quadratic equations and inequalities — Part c
• SL 2.5: Composite functions \( (f \circ g)(x) = f(g(x)) \) — Part d(i)
• SL 2.10: Solving equations… Applications of graphing skills and solving equations that relate to real-life— Part d(ii)
• SL 1.5: Introduction to logarithms with base 10 and \( e \)— Part d
▶️ Answer/Explanation

(a) Vertex form:
Expand the function:
\( f(x) = 5(x^2 + 4x + 3) = 5x^2 + 20x + 15 \).
Complete the square:
\( 5(x^2 + 4x) + 15 \)
\( = 5[(x+2)^2 – 4] + 15 \)
\( = 5(x+2)^2 – 20 + 15 \)
\( = 5(x+2)^2 – 5 \).
\( \boxed{a = 5, h = -2, k = -5} \).

(b) Sketch:

• Vertex: \( (-2, -5) \)
• x-intercepts: Set \( f(x) = 0 \implies (x+1)(x+3) = 0 \), so \( x = -1, x = -3 \).
• y-intercept: Set \( x = 0 \implies f(0) = 5(1)(3) = 15 \).
• Shape: A parabola opening upwards.

(c) Solve \( f(x) \leq 40 \):
\( 5x^2 + 20x + 15 \leq 40 \)
\( 5x^2 + 20x – 25 \leq 0 \)
Divide by 5: \( x^2 + 4x – 5 \leq 0 \)
Factor: \( (x + 5)(x – 1) \leq 0 \).
Critical values: \( x = -5, 1 \).
Solution: \( \boxed{-5 \leq x \leq 1} \).

(d)(i) Expression for \( (f \circ g)(x) \):
Substitute \( g(x) = \ln x \) into \( f(x) \):
\( \boxed{(f \circ g)(x) = 5(\ln x + 1)(\ln x + 3)} \) or \( 5(\ln x + 2)^2 – 5 \).

(d)(ii) Solve \( (f \circ g)(x) \leq 40 \):
Using the result from part (c), where the input of \( f \) must be between \( -5 \) and \( 1 \):
\( -5 \leq \ln x \leq 1 \).
Apply the base \( e \) to all parts:
\( e^{-5} \leq x \leq e^1 \).
\( \boxed{e^{-5} \leq x \leq e} \).

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