3.2Exploring Quadratic Models
Graph parabolas in standard and vertex form, find the vertex by completing the square, solve quadratic equations, and use the discriminant to determine the nature of roots.
Quadratic functions model projectile motion, profit maximization, and area optimization. The vertex form and quadratic formula are among the most used tools in all of algebra and calculus.
Essential Question
How do the parameters a, h, and k in f(x) = a(x−h)²+k determine the shape, direction, and position of a parabola — and how can we use this to solve real-world optimization problems?
Lesson Overview
A quadratic function is any function of the form f(x) = ax²+bx+c where a ≠ 0. Its graph is a U-shaped curve called a parabola. The two most useful forms are standard form and vertex form. Understanding how to move between them — and how to read key features from each — is the foundation for solving optimization problems in physics, economics, and engineering.
Standard Form
f(x) = ax² + bx + c
- y-intercept is (0, c)
- Vertex x = −b/(2a)
- Easy to read a, b, c directly
Vertex Form
f(x) = a(x−h)² + k
- Vertex is (h, k)
- Axis of symmetry: x = h
- Easy to graph and optimize
Parabola Orientations
Discriminant: b²−4ac
Completing the Square: x²+6x → (x+3)²−9
Worked Examples
Find the vertex and axis of symmetry of f(x) = 2x² − 8x + 3.
Identify a = 2, b = −8, c = 3.
Vertex x-coordinate: h = −b/(2a) = −(−8)/(2·2) = 8/4 = 2
Vertex y-coordinate: f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5
Vertex: (2, −5)
Axis of symmetry: x = 2
Convert f(x) = x² + 6x + 2 to vertex form by completing the square.
Group: f(x) = (x² + 6x) + 2
Half of 6 is 3; 3² = 9. Add and subtract 9 inside:
f(x) = (x² + 6x + 9) − 9 + 2
f(x) = (x + 3)² − 7
Solve 2x² − 3x − 2 = 0 using the quadratic formula.
a = 2, b = −3, c = −2
Discriminant: b²−4ac = 9 + 16 = 25
x = (3 ± √25) / 4 = (3 ± 5) / 4
x = 8/4 = 2 or x = −2/4 = −1/2
Determine the number and type of roots of f(x) = x² − 4x + 7.
a = 1, b = −4, c = 7
Discriminant: b²−4ac = 16 − 28 = −12
Since −12 < 0, there are 2 complex (non-real) roots.
A ball is thrown upward. Its height is h(t) = −16t² + 64t + 5 (feet). Find the maximum height and when it occurs.
This is a downward parabola (a = −16 < 0), so the vertex is the maximum.
t at vertex: t = −b/(2a) = −64/(2·−16) = −64/−32 = 2 seconds
Max height: h(2) = −16(4) + 64(2) + 5 = −64 + 128 + 5 = 69 feet
Guided Practice
Find the vertex of f(x) = 3x² + 12x − 1.
Hint: Use h = −b/(2a) first, then substitute back to find k.
Convert g(x) = x² − 10x + 18 to vertex form.
Hint: Complete the square: take half of −10, square it, add and subtract inside.
Solve x² + 5x + 6 = 0 by factoring.
Hint: Find two numbers that multiply to 6 and add to 5.
Use the discriminant to classify the roots of f(x) = 4x² − 4x + 1.
Hint: Compute b²−4ac and compare to 0.
A farmer wants to fence a rectangular area against a barn wall using 80 m of fencing for three sides. Write a quadratic function for the area and find the maximum area.
Hint: Let width = x. Then length = 80 − 2x. Area = x(80 − 2x). Find the vertex.
Key Vocabulary
Quadratic Function
A function of the form f(x) = ax²+bx+c where a ≠ 0. Its graph is a parabola.
Example: f(x) = 2x² − 3x + 1
Vertex
The turning point of a parabola — the minimum if a > 0, the maximum if a < 0.
Example: f(x) = (x−2)²+3 has vertex (2, 3)
Axis of Symmetry
The vertical line x = h that divides the parabola into two mirror-image halves.
Example: x = −b/(2a)
Vertex Form
f(x) = a(x−h)²+k, where (h, k) is the vertex and a determines direction/width.
Example: f(x) = −3(x+1)²+5
Completing the Square
Algebraic technique to rewrite ax²+bx+c in vertex form by creating a perfect square trinomial.
Example: x²+6x+9 = (x+3)²
Discriminant
The expression b²−4ac under the radical in the quadratic formula; determines the number and type of roots.
Example: If b²−4ac = 0, one real root
Quadratic Formula
x = (−b ± √(b²−4ac)) / (2a) — gives the roots of any quadratic equation ax²+bx+c = 0.
Example: x² − x − 6 = 0 → x = 3 or x = −2
Parabola
The U-shaped curve that is the graph of every quadratic function.
Example: y = x² is the simplest parabola
Check Your Understanding
Interactive Practice — 5 Questions
What is the vertex of f(x) = 2x² − 8x + 3?
Which is the vertex form of f(x) = x² + 6x + 2?
For f(x) = x² − 4x + 7, what does the discriminant tell us?
Solve 2x² − 3x − 2 = 0 using the quadratic formula.
A ball follows h(t) = −16t² + 64t + 5. When does it reach maximum height?
Practice Problems
Independent Practice
Find the vertex and axis of symmetry of f(x) = −x² + 4x − 1.
Convert f(x) = x² − 8x + 10 to vertex form by completing the square.
Solve x² − 7x + 12 = 0 by factoring.
Use the quadratic formula to solve 3x² + 5x − 2 = 0.
A rectangle has perimeter 40 cm. Write a quadratic function for its area in terms of one side length, then find the maximum area.
Common Mistakes
Writing the vertex as (b/2a, k) instead of (−b/2a, k) — forgetting the negative sign.
The vertex x-coordinate is h = −b/(2a). The negative sign is essential.
In vertex form f(x) = a(x−h)²+k, reading the vertex as (−h, k) when h is already negative.
If f(x) = (x+3)²−7, rewrite as (x−(−3))²−7, so the vertex is (−3, −7).
Forgetting to subtract the added constant when completing the square.
When you add (b/2)² inside the parentheses, you must subtract it outside to keep the equation balanced.
Concluding there are no roots when the discriminant is negative.
A negative discriminant means 2 complex (non-real) roots — the parabola simply does not cross the x-axis.
Math Tips
The axis of symmetry x = −b/(2a) is the midpoint between the two x-intercepts whenever they exist.
To quickly check vertex form, expand a(x−h)²+k and verify you get the original standard form.
For optimization word problems, the vertex always gives the maximum (a < 0) or minimum (a > 0) value.
Factoring is fastest when the discriminant is a perfect square. Otherwise, use the quadratic formula.
The y-intercept is always (0, c) in standard form — just substitute x = 0.