Unit 8 · Lesson 8.1

8.1Introduction to Quadratic Functions

Identify quadratic functions, understand the U-shaped parabola, and explore key features including the vertex, axis of symmetry, direction of opening, and intercepts.

Why This Matters

Quadratic functions model projectile motion, the shape of satellite dishes, and the path of a thrown ball. They're a major topic in Algebra 2, Precalculus, and AP Physics — and one of the most tested concepts on the SAT.

Workbook

Lesson, vocabulary, worked examples, and practice problems.

Essential Question

What makes a function quadratic, and what key features describe the shape and behavior of its graph?

Lesson Overview

A quadratic function is any function that can be written in the form f(x) = ax² + bx + c, where a ≠ 0. The graph of every quadratic function is a smooth, symmetric curve called a parabola. The parabola opens upward when a > 0 (minimum point) and downward when a < 0 (maximum point). The highest or lowest point on the parabola is called the vertex, and the vertical line through the vertex is the axis of symmetry. Understanding these features is the foundation for everything in Unit 8.

Worked Examples

Example 1

For f(x) = 2x² − 8x + 3, identify: (a) direction of opening, (b) axis of symmetry, (c) vertex, (d) y-intercept.

Identify a = 2, b = −8, c = 3.

(a) a = 2 > 0, so the parabola opens upward.

(b) Axis of symmetry: x = −b/(2a) = −(−8)/(2·2) = 8/4 = 2. So x = 2.

(c) Vertex x = 2. Find y: f(2) = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5. Vertex: (2, −5).

(d) y-intercept: set x = 0 → f(0) = 3. y-intercept: (0, 3).

Answer:Opens upward. Axis of symmetry: x = 2. Vertex: (2, −5). y-intercept: (0, 3).
Example 2

For f(x) = −x² + 4x − 1, identify the vertex and state whether it is a maximum or minimum.

Identify a = −1, b = 4, c = −1.

a = −1 < 0, so the parabola opens downward → vertex is a maximum.

Axis of symmetry: x = −b/(2a) = −4/(2·(−1)) = −4/(−2) = 2.

Vertex y: f(2) = −(2)² + 4(2) − 1 = −4 + 8 − 1 = 3.

Vertex: (2, 3). This is the maximum point.

Answer:Vertex: (2, 3). It is a maximum because a < 0 (parabola opens downward).
Example 3

Determine whether each function is quadratic. If yes, identify a, b, and c.\n(a) f(x) = 3x² − 5x + 2\n(b) g(x) = 4x − 7\n(c) h(x) = −x² + 9

(a) f(x) = 3x² − 5x + 2: highest power is 2, a = 3 ≠ 0 → quadratic. a = 3, b = −5, c = 2.

(b) g(x) = 4x − 7: highest power is 1 → linear, NOT quadratic.

(c) h(x) = −x² + 9: highest power is 2, a = −1 ≠ 0 → quadratic. a = −1, b = 0, c = 9.

Answer:(a) Quadratic: a = 3, b = −5, c = 2. (b) Not quadratic (linear). (c) Quadratic: a = −1, b = 0, c = 9.
Example 4

For f(x) = x² − 6x + 8, find the axis of symmetry, vertex, and y-intercept. Does the parabola open up or down?

a = 1, b = −6, c = 8.

Direction: a = 1 > 0 → opens upward.

Axis of symmetry: x = −(−6)/(2·1) = 6/2 = 3.

Vertex: f(3) = (3)² − 6(3) + 8 = 9 − 18 + 8 = −1. Vertex: (3, −1).

y-intercept: f(0) = 8. Point: (0, 8).

Answer:Opens upward. Axis of symmetry: x = 3. Vertex: (3, −1) — minimum. y-intercept: (0, 8).
Example 5

A ball is thrown upward. Its height in feet is modeled by h(t) = −16t² + 64t + 5, where t is time in seconds. What is the maximum height and when does it occur?

a = −16, b = 64, c = 5. Since a < 0, the parabola opens downward → vertex is the maximum.

Time of maximum: t = −b/(2a) = −64/(2·(−16)) = −64/(−32) = 2 seconds.

Maximum height: h(2) = −16(2)² + 64(2) + 5 = −64 + 128 + 5 = 69 feet.

Answer:Maximum height of 69 feet occurs at t = 2 seconds.

Guided Practice

Guided Practice Video: Introduction to Quadratic Functions

Review the standard form of a quadratic function, the parabola, and key features before completing the guided problems below.

Video by Sang Real Math

Watch on YouTube ↗
Guided Problem 1

For f(x) = x² + 4x − 5, find the axis of symmetry and vertex.

Hint: Use x = −b/(2a) with a = 1, b = 4. Then substitute x back into f(x).

Guided Problem 2

For f(x) = −3x² + 12x − 7, does the parabola open up or down? Is the vertex a max or min?

Hint: Look at the sign of a. Negative a → opens down → vertex is a maximum.

Guided Problem 3

Find the y-intercept of f(x) = 5x² − 3x + 11.

Hint: Set x = 0. The y-intercept is always c in standard form.

Guided Problem 4

Is f(x) = 7x − 2 a quadratic function? Explain.

Hint: Check the highest power of x. Quadratic functions must have an x² term with a non-zero coefficient.

Guided Problem 5

For f(x) = −2x² + 8x − 3, find the vertex and state the maximum value of the function.

Hint: Find the axis of symmetry first, then substitute to find the vertex y-value. That y-value is the maximum.

Key Vocabulary

Quadratic Function

A function of the form f(x) = ax² + bx + c, where a ≠ 0. The highest power of x is 2.

Parabola

The U-shaped graph of a quadratic function. Every parabola has a vertex and an axis of symmetry.

Vertex

The highest or lowest point on a parabola. It is the point where the parabola changes direction.

Axis of Symmetry

The vertical line x = h that passes through the vertex and divides the parabola into two mirror-image halves.

Direction of Opening

Whether the parabola opens upward (a > 0) or downward (a < 0).

Minimum

The lowest point of a parabola that opens upward. The y-value of the vertex is the minimum value.

Maximum

The highest point of a parabola that opens downward. The y-value of the vertex is the maximum value.

x-intercept(s)

The point(s) where the parabola crosses the x-axis. Also called roots or zeros of the quadratic.

y-intercept

The point where the parabola crosses the y-axis. Found by evaluating f(0) = c.

Standard Form

f(x) = ax² + bx + c, where a, b, c are real numbers and a ≠ 0.

Leading Coefficient

The value of a in f(x) = ax² + bx + c. Determines direction of opening and width of the parabola.

Zero of a Function

An x-value where f(x) = 0. Zeros are the x-intercepts of the graph.

Practice Questions

Interactive Practice — 5 Questions

1

Which of the following is a quadratic function?

2

For f(x) = 3x² − 12x + 5, what is the axis of symmetry?

3

A parabola opens downward. Which statement is true about its vertex?

4

What is the y-intercept of f(x) = −2x² + 5x − 8?

5

For f(x) = x² − 4x + 3, what is the vertex?

Independent Practice

Independent Practice

1

For f(x) = x² − 2x − 8, find the axis of symmetry and vertex.

2

For f(x) = −x² + 6x − 5, find the axis of symmetry and vertex.

3

Identify a, b, and c for f(x) = 3x² + 0x − 4. Does it open up or down?

4

Find the y-intercept of f(x) = −2x² + 7x − 3.

5

Is f(x) = x³ − 2x + 1 a quadratic function? Explain.

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Common Mistakes

Confusing the axis of symmetry formula — writing x = b/2a instead of x = −b/(2a).

The axis of symmetry is x = −b/(2a). The negative sign is essential.

Mixing up minimum and maximum — saying a parabola opening upward has a maximum.

Opens up (a > 0) → minimum at vertex. Opens down (a < 0) → maximum at vertex.

Thinking the vertex is always at the origin.

The vertex is at (h, k) where h = −b/(2a) and k = f(h). It moves based on the equation.

Confusing x-intercepts (zeros/roots) with the y-intercept.

x-intercepts: set y = 0 and solve. y-intercept: set x = 0 and evaluate. They are different points.

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Math Tips

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Sign of a determines direction: positive a → opens up (smile shape), negative a → opens down (frown shape).

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Axis of symmetry formula: x = −b/(2a). Memorize this — you will use it in every quadratic problem.

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Always substitute the axis of symmetry x-value back into f(x) to find the vertex y-coordinate. Do not skip this step.

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SAT/ACT tip: quadratic functions are heavily tested. Know how to find the vertex, axis of symmetry, and direction of opening quickly.