Welcome to the Quadratic Detective Agency!
In this chapter, we are learning a crucial skill: how to work backward. Instead of starting with an equation and sketching the graph, we are given key features of the parabola (the shape of a quadratic graph) and must construct the formula that created it.
This skill is essential in Modelling, allowing mathematicians and engineers to find the precise path of a projectile, the curve of a satellite dish, or the optimal shape of an arch bridge.
Quick Review: The Quadratic Function
A quadratic function is any function that can be written in the general form:
Standard/General Form: \(y = ax^2 + bx + c\)
While this form is useful for calculating the y-intercept (\(c\)), it is usually not the easiest form to use when finding the equation from given points. For that, we rely on two more specialized forms.
1. Our Essential Toolkit: Specialized Forms
To find a quadratic function based on geometric information (like the vertex or the intercepts), we must choose the correct specialized form.
Tool 1: The Vertex Form (or Turning Point Form)
This form is perfect when you know the highest or lowest point of the parabola.
Formula: \(y = a(x - h)^2 + k\)
- \((h, k)\) is the vertex (the turning point).
- \(a\) determines the direction (up if \(a > 0\), down if \(a < 0\)) and the vertical stretch/compression of the parabola.
- Memory Aid: Remember the sign change inside the bracket! If the vertex is at \((2, 5)\), the equation starts with \((x - 2)^2\).
Tool 2: The Intercept Form (or Factored Form)
This form is perfect when you know where the graph crosses the x-axis (the roots/zeros).
Formula: \(y = a(x - p)(x - q)\)
- \(p\) and \(q\) are the x-intercepts (the roots or zeros). These are the points where \(y = 0\).
- \(a\) again determines the stretch and direction.
- Common Mistake Alert: If the intercepts are \(x = -3\) and \(x = 4\), the factors are \((x + 3)\) and \((x - 4)\). The signs are always opposite!
2. Case A: Finding the Function from the Vertex and One Point
This is one of the most common types of problems. You are given the exact turning point and one other random point the parabola passes through.
Step-by-Step Process (Using the Vertex Form)
Example: Find the equation of the quadratic whose vertex is \((3, 1)\) and which passes through the point \((5, 9)\).
Step 1: Identify \(h\) and \(k\) and Substitute into the Vertex Form.
The vertex \((h, k)\) is \((3, 1)\).
Start with: \(y = a(x - h)^2 + k\)
Substitution: \(y = a(x - 3)^2 + 1\)
Step 2: Use the Other Point \((x, y)\) to Solve for \(a\).
The other point \((5, 9)\) means \(x = 5\) and \(y = 9\). Substitute these values into the equation from Step 1.
\(9 = a(5 - 3)^2 + 1\)
\(9 = a(2)^2 + 1\)
\(9 = 4a + 1\)
Step 3: Isolate \(a\).
\(9 - 1 = 4a\)
\(8 = 4a\)
\(a = 2\)
Step 4: Write the Final Equation.
Substitute the calculated value of \(a\) back into the Vertex Form (from Step 1).
Final Answer: \(y = 2(x - 3)^2 + 1\)
Key Takeaway for Case A:
Start with the Vertex Form \(y = a(x-h)^2 + k\). The vertex gives you \(h\) and \(k\) immediately. The extra point provides the \(x\) and \(y\) needed to calculate the stretch factor \(a\).
3. Case B: Finding the Function from x-Intercepts and One Point
If you know where the graph crosses the x-axis, the Intercept Form is your best friend.
Step-by-Step Process (Using the Intercept Form)
Example: Find the equation of the quadratic that has x-intercepts at \(x = -2\) and \(x = 4\), and which passes through the point \((1, -9)\).
Step 1: Identify \(p\) and \(q\) and Substitute into the Intercept Form.
The intercepts are \(p = -2\) and \(q = 4\).
Start with: \(y = a(x - p)(x - q)\)
Substitution (remember opposite signs in the factor!):
\(y = a(x - (-2))(x - 4)\)
\(y = a(x + 2)(x - 4)\)
Step 2: Use the Other Point \((x, y)\) to Solve for \(a\).
The point \((1, -9)\) means \(x = 1\) and \(y = -9\).
\(-9 = a(1 + 2)(1 - 4)\)
\(-9 = a(3)(-3)\)
\(-9 = -9a\)
Step 3: Isolate \(a\).
\(a = 1\)
Step 4: Write the Final Equation (and Expand if requested).
Substitute \(a = 1\):
\(y = 1(x + 2)(x - 4)\)
If the question asks for the standard form \(y = ax^2 + bx + c\), you must expand the brackets:
\(y = x^2 - 4x + 2x - 8\)
Final Answer: \(y = x^2 - 2x - 8\)
Did you know? Finding the Vertex from Intercepts
If you are given the intercepts, you can easily find the x-coordinate of the vertex!
A parabola is always symmetrical about its vertex. The x-coordinate of the vertex is exactly halfway between the intercepts.
If intercepts are \(p\) and \(q\), the x-coordinate of the vertex is: \(x_{vertex} = \frac{p + q}{2}\).
In the example above (\(p = -2, q = 4\)): \(x_{vertex} = \frac{-2 + 4}{2} = 1\). This matches the \(x\) value of the point \((1, -9)\), confirming that \((1, -9)\) was the vertex!
4. The Special Case: When \(a = 1\) (Syllabus E3.4c)
Sometimes, the exam simplifies the problem by telling you the stretch factor \(a\) is 1. If \(a=1\), the parabola is identical in shape to \(y = x^2\), just shifted and positioned differently.
Scenario 1: Given Vertex \((h, k)\) and \(a=1\)
If the vertex is \((-1, 5)\) and \(a=1\), you do not need a second point! You can write the equation straight away:
\(y = 1(x - (-1))^2 + 5\)
Equation: \(y = (x + 1)^2 + 5\)
Scenario 2: Given x-intercepts \(p\) and \(q\) and \(a=1\)
If the x-intercepts are \(x=0\) and \(x=6\), and \(a=1\):
\(y = 1(x - 0)(x - 6)\)
Equation: \(y = x(x - 6)\) or \(y = x^2 - 6x\)
Tip for struggling students: If the question is simple and only gives the vertex OR the intercepts (but not both), assume they are testing the \(a=1\) case, or look very closely for a statement that gives you the value of \(a\). If you are asked to find \(a\) yourself, you must be given three pieces of information (like two points and a vertex, or two points and two intercepts).
5. Final Summary: Choosing the Right Starting Point
Quick Review Box:
- If given the Vertex: Use \(y = a(x - h)^2 + k\)
- If given the x-Intercepts: Use \(y = a(x - p)(x - q)\)
- ALWAYS: Use the extra point \((x, y)\) to find the unknown value of \(a\).
Don't worry if this seems tricky at first! Quadratic functions have many forms, but once you match the information you have (vertex or intercepts) to the correct formula, the process becomes simple algebra to find \(a\). Practice makes perfect in choosing the right tool for the job!