8.2Graphing Quadratic Functions
Graph parabolas from standard form by locating the vertex, axis of symmetry, and intercepts — then use symmetry to plot a complete, accurate curve.
Why This Matters
Graphing parabolas builds visual intuition for how quadratic relationships behave. You'll use this skill in Algebra 2 when analyzing transformations, in Physics when modeling projectile paths, and in Calculus when finding maxima and minima.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How do the values of a, b, and c in y = ax² + bx + c determine the shape, direction, and position of a parabola?
Lesson Overview
Every quadratic function y = ax² + bx + c produces a U-shaped curve called a parabola. The key to graphing it accurately is finding the vertex — the turning point — using the axis of symmetry formula x = −b/(2a). Once you have the vertex, you can build a table of values on one side and mirror those points across the axis to complete the graph. The sign of a tells you whether the parabola opens up (minimum vertex) or down (maximum vertex), and the size of |a| controls how narrow or wide the curve is.
Key Concepts & Visuals
Anatomy of a Parabola — y = x² − 4
Vertex (0,−4) · Axis of symmetry x=0 · x-intercepts (±2, 0)
Finding the Vertex from Standard Form y = ax² + bx + c
Axis of symmetry
Use the formula x = −b / (2a)
y = 2x² − 8x + 3 → x = −(−8)/(2·2) = 8/4 = 2
Vertex x-coordinate
The axis of symmetry IS the x-coordinate of the vertex.
x = 2
Vertex y-coordinate
Substitute x back into the original equation.
y = 2(2)²−8(2)+3 = 8−16+3 = −5
Write the vertex
Vertex = (x, y)
Vertex = (2, −5)
Table of Values — y = 2x² − 8x + 3
| x | y = 2x²−8x+3 | Point | Note |
|---|---|---|---|
| -2 | 27 | (-2, 27) | |
| -1 | 13 | (-1, 13) | |
| 0 | 3 | (0, 3) | y-intercept |
| 1 | -3 | (1, -3) | |
| 2 | -5 | (2, -5) | ★ Vertex |
| 3 | -3 | (3, -3) | |
| 4 | 3 | (4, 3) |
Vertex at (2, −5). Parabola opens upward (a = 2 > 0).
Opens Up vs. Opens Down — Effect of a
a > 0 → Opens Upward ∪
y = x², y = 2x², y = 0.5x²
- Vertex is the minimum point
- Parabola opens upward
- Larger |a| → narrower parabola
a < 0 → Opens Downward ∩
y = −x², y = −3x², y = −0.5x²
- Vertex is the maximum point
- Parabola opens downward
- Larger |a| → narrower parabola
Graphing a Quadratic — Step-by-Step Checklist
Worked Examples
Find the axis of symmetry and vertex of y = x² − 4x + 3
Identify: a=1, b=−4, c=3
Axis of symmetry: x = −(−4)/(2·1) = 4/2 = 2
Vertex y: y = (2)²−4(2)+3 = 4−8+3 = −1
Vertex = (2, −1)
Find the axis of symmetry and vertex of y = −2x² + 8x − 5
Identify: a=−2, b=8, c=−5
Axis of symmetry: x = −8/(2·(−2)) = −8/(−4) = 2
Vertex y: y = −2(4)+8(2)−5 = −8+16−5 = 3
Vertex = (2, 3); opens downward (a<0) → maximum
Find the y-intercept of y = 3x² − 5x + 7
Set x = 0: y = 3(0)²−5(0)+7 = 7
y-intercept = (0, 7)
In standard form, the y-intercept is always (0, c)
Build a table of values for y = x² − 2x − 3 and identify the vertex
Axis: x = −(−2)/(2·1) = 1; vertex x = 1
x=−1: y=1+2−3=0; x=0: y=0−0−3=−3; x=1: y=1−2−3=−4 (vertex)
x=2: y=4−4−3=−3; x=3: y=9−6−3=0
Vertex = (1, −4); x-intercepts at (−1,0) and (3,0)
Describe the parabola y = −3x² + 6x − 2 without graphing
a = −3 < 0 → opens downward, maximum vertex
|a| = 3 > 1 → narrower than y = x²
Axis: x = −6/(2·(−3)) = −6/(−6) = 1
Vertex y: y = −3(1)+6(1)−2 = −3+6−2 = 1 → Vertex (1, 1)
y-intercept: (0, −2)
Use symmetry: y = x² − 6x + 5. Given point (0, 5), find its mirror.
Axis of symmetry: x = 6/2 = 3
Point (0, 5) is 3 units left of axis (3−0=3)
Mirror point is 3 units right of axis: x = 3+3 = 6
Verify: y(6) = 36−36+5 = 5 → Mirror point: (6, 5) ✓
Guided Practice
Guided Practice Video: Graphing Quadratic Functions
Review graphing parabolas using vertex form and standard form before completing the guided problems below.
Video by Sang Real Math
Watch on YouTube ↗Find the axis of symmetry of y = x² + 6x + 8.
Hint: Use x = −b/(2a). Identify a=1 and b=6.
Find the vertex of y = x² + 6x + 8.
Hint: Substitute the axis of symmetry x-value into the equation to find y.
Does y = −4x² + 2x − 1 open upward or downward? Is the vertex a max or min?
Hint: Look at the sign of a. Negative a → opens downward → maximum vertex.
Find the y-intercept of y = 5x² − 3x + 11.
Hint: Set x = 0. In standard form, the y-intercept is always (0, c).
Build a table of values for y = x² − 4 using x = −3, −2, −1, 0, 1, 2, 3.
Hint: Substitute each x-value. The vertex is at x=0 (since b=0). Which points are mirrors of each other?
The axis of symmetry of a parabola is x = 4. One point on the parabola is (1, 7). Find its mirror point.
Hint: The point is 3 units left of the axis (4−1=3). The mirror is 3 units right: x = 4+3 = 7.
Key Vocabulary
Axis of Symmetry
The vertical line x = −b/(2a) that divides the parabola into two mirror-image halves.
Vertex
The point where the parabola changes direction. The minimum (a>0) or maximum (a<0) of the function.
y-intercept
The point where the parabola crosses the y-axis. Found by setting x = 0; equals (0, c) in standard form.
x-intercepts
Points where the parabola crosses the x-axis (y = 0). Also called roots or zeros. A parabola may have 0, 1, or 2.
Table of Values
A chart of (x, y) pairs generated by substituting x-values into the equation. Used to plot accurate points.
Symmetry
Every point on one side of the axis of symmetry has a mirror-image point at the same distance on the other side.
Practice Questions
Interactive Practice — 5 Questions
What is the axis of symmetry of y = x² − 6x + 5?
What is the vertex of y = x² − 6x + 5?
For y = −2x² + 8x − 3, does the parabola open up or down?
What is the y-intercept of y = 4x² − 3x + 7?
A parabola has axis of symmetry x = 2. One point is (0, 5). What is its mirror point?
Independent Practice
Independent Practice
Find the axis of symmetry: y = x² − 8x + 12
Find the vertex: y = x² − 8x + 12
Find the axis of symmetry and vertex: y = −x² + 4x − 3
Find the y-intercept: y = 2x² − 7x + 4
Does y = 3x² − 6x + 1 open up or down? Max or min?
Common Mistakes
Plotting the vertex at (b/2a, …) instead of (−b/2a, …).
The axis of symmetry is x = −b/(2a). Always include the negative sign.
Forgetting to find and plot the y-intercept (0, c) as a reference point.
The y-intercept is always at (0, c) in standard form. Plot it first — it's the easiest point to find.
Drawing only one side of the parabola without reflecting across the axis of symmetry.
A parabola is symmetric. For every point on one side, there's a mirror image on the other side of the axis.
Confusing the vertex with the y-intercept when a = 1, b = 0 — assuming the vertex is at the origin.
The vertex is at (−b/2a, f(−b/2a)). When b = 0, the axis is x = 0, but the vertex height depends on c.
Math Tips
Axis of symmetry first — x = −b/(2a) gives you the x-coordinate of the vertex for free.
Always substitute the axis x-value back into the equation to find the vertex y-coordinate — never use c.
y-intercept is always (0, c) in standard form — no calculation needed.
Use symmetry: once you have points on one side of the axis, mirror them to the other side.