Algebra 2

Completing the Square & Quadratic Formula

You’ve seen quadratics in standard form: ax^2 + bx + c. But there’s another form that reveals the vertex directly — and the technique to get there is called completing the square. It’s also how the quadratic formula is derived.

Part 1: Standard Form — What Do a, b, c Control?

Start with the standard form and explore how each coefficient shapes the parabola:

a (curvature)1

-33

b (tilt)-2

-66

c (y-intercept)-3

-55

f(x) = 1x^2 + -2x + -3

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Notice how the coefficients work:

a controls the curvature: positive = opens up, negative = opens down
b shifts the vertex left and right (and up/down in a complex way)
c is where the parabola crosses the y-axis (set x = 0)

But finding the vertex from a, b, c isn’t obvious. That’s why we need vertex form!

Part 2: Vertex Form — The Clear Picture

The vertex form of a quadratic is:

f(x) = a(x - h)^2 + k

Here, (h, k) is the vertex — the highest or lowest point. Let’s see it:

a (curvature)1

-33

h (vertex x)1

-55

k (vertex y)-4

-55

f(x) = 1(x - 1)^2 + -4

Try This

Now the vertex is obvious! (h, k) is right there in the equation.

h slides the parabola left and right
k slides it up and down
a still controls the curvature

Compare this to standard form — the vertex is hidden in a, b, c but visible in h, k.

Part 3: Completing the Square — The Transformation

Here’s the step-by-step process. Starting from ax^2 + bx + c:

Factor out a from the first two terms: a(x^2 + (b/a)x) + c
Add and subtract (b/2a)^2 inside the parentheses
Factor the perfect square trinomial
Simplify to get a(x - h)^2 + k

The vertex is at:

h = -\frac00, \quad k = c - \frac{b^2}0

Let’s verify this with our sliders. Set a, b, c and see the computed vertex:

0.53

-66

-55

\text{Standard: } 1x^2 + 4x + 1

h = -\frac{4}{2 \times 1}, \quad k = 1 - \frac{4^2}{4 \times 1}

Connection

The vertex formula h = -b/(2a) comes directly from completing the square. It’s not a separate thing to memorize — it’s a consequence of the algebra. Try a = 1, b = 4, c = 1. The vertex should be at h = -2, k = -3.

Part 4: The Discriminant — How Many Roots?

The discriminant determines how many times the parabola crosses the x-axis:

\Delta = b^2 - 4ac

0.52

-66

c-3

-55

\Delta = 2^2 - 4(1)(-3)

Try This

Experiment with the discriminant:

Delta > 0: Two real roots — the parabola crosses the x-axis twice (try a=1, b=2, c=-3)
Delta = 0: One repeated root — the vertex touches the x-axis (try a=1, b=-2, c=1)
Delta < 0: No real roots — the parabola doesn’t reach the x-axis (try a=1, b=0, c=2)

The discriminant tells you the answer before you solve!

Part 5: The Quadratic Formula

Completing the square on the general equation ax^2 + bx + c = 0 gives us the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}0

The discriminant (b^2 - 4ac) is right there under the square root — that’s why it determines the number of roots!

0.53

b-1

-66

c-6

-55

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}

Challenge

Challenge: Use the quadratic formula to solve these, then verify with the graph:

x^2 - 5x + 6 = 0 (set a=1, b=-5, c=6)
2x^2 + 3x - 2 = 0 (set a=2, b=3, c=-2)
x^2 + 2x + 5 = 0 (set a=1, b=2, c=5) — what happens?

For #3, the discriminant is negative. The quadratic formula gives complex numbers — the parabola never crosses the x-axis!

Wrapping Up

Concept	Key Idea
Standard form	ax^2 + bx + c — good for finding y-intercept
Vertex form	a(x-h)^2 + k — reveals vertex (h,k) directly
Completing the square	Transforms standard to vertex form
Vertex location	h = -b/(2a), k = c - b^2/(4a)
Discriminant	b^2 - 4ac tells you: 2 roots, 1 root, or 0 real roots
Quadratic formula	x = (-b +/- sqrt(discriminant)) / (2a)

Completing the square isn’t just a technique — it’s the reason the quadratic formula works. Every time you use the formula, you’re doing completing the square in disguise.

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