IB Mathematics AA SL The quadratic function Study Notes
IB Mathematics AA SL The quadratic function Study Notes Offer a clear explanation of The quadratic function , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic The quadratic function
The Quadratic Function: Quadratic Equations
A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable or unknown. Quadratic equations represent parabolic curves when graphed on the coordinate plane.
1. Solving by Factorizing
Factorizing is one of the simplest methods for solving quadratic equations. The goal is to express the quadratic equation as a product of two binomials. Once factorized, you can set each factor equal to zero and solve for \( x \).
- Steps:
- Make the equation equal to zero (general form).
- Factor the quadratic equation: Think about how the equation can become zero.
- Write the solution as \( x = \_, \_ \).
2. Solving by Using the Quadratic Formula
The quadratic formula is a general solution to any quadratic equation. It is derived from completing the square on the general form of the quadratic equation.
The quadratic formula is:
\(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)
- Where:
- \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
- The symbol \( \pm \) means there are two possible solutions: one with the positive square root and one with the negative square root.
- The term under the square root, \( b^2 – 4ac \), is called the discriminant, and it indicates the nature of the roots.
3. Solving Using TI-84
The TI-84 calculator provides a straightforward way to solve quadratic equations, especially when factorization is difficult. It uses the quadratic formula automatically when the equation is inputted into the Poly Root Finder app.
Example: Solve \( 5x^2 – 8x + 12 = 0 \)
- Press APPS.
- Select PlySmit 2 and then Poly Root Finder.
- Enter the values for \( a = 5 \), \( b = -8 \), and \( c = 12 \).
- The calculator will return the solutions for \( x \).
4. Solving by Completing the Square (Vertex Form)
Completing the square is a method that transforms a quadratic equation into a perfect square trinomial. This is especially useful when you want to write the equation in vertex form, which is ideal for graphing and finding translations.
The vertex form of a quadratic equation is:
\(y = a(x – h)^2 + k \)
Where \( (h, k) \) represents the vertex of the parabola.
Example:
- Start with the equation \( x^2 + 10x \).
- To complete the square, add and subtract \( 25 \):
- \( (x^2 + 10x + 25) – 25 = (x + 5)^2 – 25 \)
- The resulting equation is \( (x + 5)^2 – 25 \), which is in vertex form.
Quadratic Forms
Quadratic equations can be expressed in different forms. Each form provides different insights into the equation, particularly when analyzing the graph of the quadratic function.
- 1. General Form: \( ax^2 + bx + c = 0 \)
- This form is commonly used in calculations with graphing calculators (GDC).
- The axis of symmetry formula is \( x = \frac{-b}{2a} \). This formula gives the x-coordinate of the vertex.
- The y-intercept is the constant \( c \), the value where the graph crosses the y-axis.
- 2. Factored Form: \( a(x + p)(x + q) = 0 \)
- In this form, the x-intercepts are easily identified as \( x = -p \) and \( x = -q \).
- 3. Vertex Form: \( a(x – h)^2 + k = 0 \)
- This form is useful for identifying the vertex of the parabola, which is \( (h, k) \).
- Graph transformations, such as shifts and stretches, are easier to identify in this form.
Discriminant
The discriminant is the expression \( b^2 – 4ac \) that appears under the square root in the quadratic formula. It provides important information about the nature of the roots of the quadratic equation.
- If \( b^2 – 4ac > 0 \): There are 2 real roots (distinct solutions).
- If \( b^2 – 4ac = 0 \): There is 1 real root (a repeated root).
- If \( b^2 – 4ac < 0 \): There are no real roots, only complex (imaginary) roots.
In this context, “roots” refer to the x-intercepts of the parabola, or the solutions to the quadratic equation.
Examples of Quadratic Functions
Below are some examples demonstrating the different methods for solving quadratic equations:
Example 1: Solving by Factorizing
Solve the quadratic equation: \( x^2 – 5x + 6 = 0 \)
Step 1: Factorize the equation.
We can factorize this equation as: \( (x – 2)(x – 3) = 0 \)
Step 2: Set each factor equal to zero:
\( x – 2 = 0 \quad \text{or} \quad x – 3 = 0 \)
Solution: \( x = 2 \quad \text{or} \quad x = 3 \)
Example 2: Solving by Using the Quadratic Formula
Solve the quadratic equation: \( 2x^2 – 4x – 6 = 0 \)
Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Here, \( a = 2, b = -4, c = -6 \).
Step 1: Substitute the values of \( a \), \( b \), and \( c \) into the formula.
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)} \)
Step 2: Simplify the equation:
\( x = \frac{4 \pm \sqrt{16 + 48}}{4} = \frac{4 \pm \sqrt{64}}{4} \)
\( x = \frac{4 \pm 8}{4} \)
Solution: \( x = 3 \quad \text{or} \quad x = -1 \)
Example 3: Solving by Completing the Square
Solve the quadratic equation: \( x^2 + 6x – 7 = 0 \)
Step 1: Move the constant term to the right side of the equation.
\( x^2 + 6x = 7 \)
Step 2: Complete the square by adding \( \left(\frac{6}{2}\right)^2 = 9 \) to both sides.
\( x^2 + 6x + 9 = 7 + 9 \)
\( (x + 3)^2 = 16 \)
Step 3: Take the square root of both sides.
\( x + 3 = \pm 4 \)
Step 4: Solve for \( x \).
Solution: \( x = 1 \quad \text{or} \quad x = -7 \)
Example 4: Solving by Using a Graphing Calculator (TI-84)
Example quadratic equation: \( 5x^2 – 8x + 12 = 0 \)
Step 1: Turn on the calculator and select the app “Poly Root Finder”.
Step 2: Enter the values for \( a = 5, b = -8, c = 12 \).
Step 3: The calculator will give you the roots: \( x = \frac{-(-8) \pm \sqrt{(-8)^2 – 4(5)(12)}}{2(5)} \)
Solution: The calculator shows the roots, which can be approximated for real-world solutions.
IB Mathematics AA SL The quadratic function Exam Style Worked Out Questions
Question
Let \(f(x) = a{(x – h)^2} + k\). The vertex of the graph of \(f\) is at \((2, 3)\) and the graph passes through \((1, 7)\).
Write down the value of \(h\) and of \(k\).
Find the value of \(a\).
▶️Answer/Explanation
Markscheme
\(h = 2,{\text{ }}k = 3\) A1A1 N2
[2 marks]
attempt to substitute \((1,7)\) in any order into their \(f(x)\) (M1)
eg \(7 = a{(1 – 2)^2} + 3{\text{, }}7 = a{(1 – 3)^2} + 2{\text{, }}1 = a{(7 – 2)^2} + 3\)
correct equation (A1)
eg \(7 = a + 3\)
a = 4 A1 N2
[3 marks]
Question
Let \(f(x) = 3{x^2} – 6x + p\). The equation \(f(x) = 0\) has two equal roots.
Write down the value of the discriminant.[2]
Hence, show that \(p = 3\).[1]
The graph of \(f\)has its vertex on the \(x\)-axis.
Find the coordinates of the vertex of the graph of \(f\).[4]
The graph of \(f\) has its vertex on the \(x\)-axis.
Write down the solution of \(f(x) = 0\).[1]
The graph of \(f\) has its vertex on the \(x\)-axis.
The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(a\).[1]
The graph of \(f\) has its vertex on the \(x\)-axis.
The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(h\).[1]
The graph of \(f\) has its vertex on the \(x\)-axis.
The function can be written in the form \(f(x) = a{(x – h)^2} + k\). Write down the value of \(k\).[1]
The graph of \(f\) has its vertex on the \(x\)-axis.
The graph of a function \(g\) is obtained from the graph of \(f\) by a reflection of \(f\) in the \(x\)-axis, followed by a translation by the vector \(\left( \begin{array}{c}0\\6\end{array} \right)\). Find \(g\), giving your answer in the form \(g(x) = A{x^2} + Bx + C\).[4]
▶️Answer/Explanation
Markscheme
correct value \(0\), or \(36 – 12p\) A2 N2
[2 marks]
correct equation which clearly leads to \(p = 3\) A1
eg \(36 – 12p = 0,{\text{ }}36 = 12p\)
\(p = 3\) AG N0
[1 mark]
METHOD 1
valid approach (M1)
eg \(x = – \frac{b}{{2a}}\)
correct working A1
eg \( – \frac{{( – 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
METHOD 2
valid approach (M1)
eg \(f(x) = 0\), factorisation, completing the square
correct working A1
eg \({x^2} – 2x + 1 = 0,{\text{ }}(3x – 3)(x – 1),{\text{ }}f(x) = 3{(x – 1)^2}\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
METHOD 3
valid approach using derivative (M1)
eg \(f'(x) = 0,{\text{ }}6x – 6\)
correct equation A1
eg \(6x – 6 = 0\)
correct answers A1A1 N2
eg \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)
[4 marks]
\(x = 1\) A1 N1
[1 mark]
\(a = 3\) A1 N1
[1 mark]
\(h = 1\) A1 N1
[1 mark]
\(k = 0\) A1 N1
[1 mark]
attempt to apply vertical reflection (M1)
eg \( – f(x),{\text{ }} – 3{(x – 1)^2}\), sketch
attempt to apply vertical shift 6 units up (M1)
eg \( – f(x) + 6\), vertex \((1, 6)\)
transformations performed correctly (in correct order) (A1)
eg \( – 3{(x – 1)^2} + 6,{\text{ }} – 3{x^2} + 6x – 3 + 6\)
\(g(x) = – 3{x^2} + 6x + 3\) A1 N3
[4 marks]