Quadratic Equations & Graphing

Published

January 20, 2026

Finding an equation from a graph is like being a math detective. You’re given clues (intercepts, vertices, special points) and you need to figure out the “formula” that created the pattern.

Scientists do this all the time: they collect data, plot it, and then find the equation that fits. That equation lets them predict the future — like predicting where a rocket will land or how a disease will spread.

Topics Covered

  • Constructing quadratic equations from graphs
  • Factored form vs completed square form
  • Using x-intercepts, y-intercepts, and symmetry
  • Extension to cubic polynomials
  • Signs of coefficients \(a\), \(b\), \(c\)

These all mean the same thing: a value of \(x\) where \(y = 0\) (the graph touches or crosses the x-axis).

For \(y = (x-3)(x+1)\), the roots are \(x = 3\) and \(x = -1\).

Lecture Video

Key Video Frames

t = 13:20

t = 17:20

t = 17:40

t = 20:40

Two Major Forms

If \(x = r_1\), the first parenthesis becomes zero, making the whole thing zero — the point is on the x-axis!

Example: Roots at \(x = -2\) and \(x = 3\): \(y = k(x + 2)(x - 3)\) Plug in \(x = -2\): \(y = k(0)(-5) = 0\) — on the x-axis!

Form Expression Best when you know…
Factored form \(y = k(x - r_1)(x - r_2)\) The roots/x-intercepts
Completed square \(y = a(x - h)^2 + k\) The vertex/AOS

Example 1: Find equation from two x-intercepts

Given: x-intercepts at \(-2\) and \(3\), y-intercept at \(2\)

  1. From roots: \(y = k(x + 2)(x - 3)\)
  2. Plug in \((0, 2)\): \(2 = k(2)(-3) = -6k \Rightarrow k = -\frac{1}{3}\)
  3. Answer: \(y = -\frac{1}{3}(x + 2)(x - 3)\)

Try it — drag the roots and see how the parabola changes:

When two points have the same y-value, they must be mirror images across the axis of symmetry!

AOS = average of their x-values (just the midpoint formula). Think of it like a balance point — the mirror is exactly in the middle.

Example 2: Using symmetry from equal y-values

Given: Points \((1, 6)\) and \((-5, 6)\) share the same \(y\)-value, plus \((0, -1)\)

  1. AOS at \(x = \frac{1 + (-5)}{2} = -2\) (midpoint of symmetric points)
  2. Treat \(y = 6\) as a shifted x-axis: \(y = k(x - 1)(x + 5) + 6\)
  3. Plug in \((0, -1)\): \(-1 = k(0-1)(0+5) + 6 \Rightarrow k = \frac{7}{5}\)

Key insight: Points with equal y-values reveal the axis of symmetry!

Feature Quadratic (\(x^2\)) Cubic (\(x^3\))
Highest power 2 3
Shape U or upside-down U S-curve
Max roots 2 3
End behavior Same direction both sides Opposite directions
Max turning points 1 2

The key insight: odd powers go in opposite directions at the extremes, even powers go in the same direction.

Example 3: Extension to cubic polynomials

Given: Roots at \(-6\), \(-2\), \(4\); y-intercept at \(-4\)

  1. From roots: \(y = k(x + 6)(x + 2)(x - 4)\)
  2. Plug in \((0, -4)\): \(-4 = k(6)(2)(-4) = -48k \Rightarrow k = \frac{1}{12}\)

Determining signs of \(a\), \(b\), \(c\)

Given a graph of \(y = ax^2 + bx + c\):

  • \(c\) = y-intercept (positive if graph crosses y-axis above origin)
  • \(a\) = opens up (\(a > 0\)) or down (\(a < 0\))
  • \(b\): use AOS formula \(x = -\frac{b}{2a}\)
    • If AOS is positive and \(a < 0\): then \(b > 0\) (two negatives cancel)

Cheat Sheet

How to find an equation from a graph:

  1. Know the roots? → Use factored form \(y = k(x - r_1)(x - r_2)\)
  2. Know the vertex? → Use vertex form \(y = a(x - h)^2 + k\)
  3. Two points with same y? → Their midpoint = axis of symmetry
  4. Always: use one extra point to solve for the unknown coefficient \(k\)
Concept Formula
Factored form \(y = k(x - r_1)(x - r_2)\)
Midpoint (symmetry) \(x_{AOS} = \frac{x_1 + x_2}{2}\)
AOS from coefficients \(x = -\frac{b}{2a}\)
Cubic factored form \(y = k(x - r_1)(x - r_2)(x - r_3)\)