Aprender · Algebra 2

Quadratic Functions Mastery: Vertex Form, Completing the Square, and the Discriminant

Algebra 1 introduced parabolas. Algebra 2 makes you fluent — switch between forms, complete the square in your head, and read the discriminant to know exactly how many real roots before solving.

10 min TEKS 4A,4B,4C,4D,4F Algebra 2

From "knowing the formula" to "fluent across forms"

Algebra 1 gave you the quadratic formula and one form. Algebra 2 expects fluency: vertex form for graphing, standard form for the formula, factored form for roots — and the ability to switch between them in seconds.

The three forms

Standard: f(x) = ax2 + bx + c Vertex: f(x) = a(x − h)2 + k Factored: f(x) = a(x − r1)(x − r2) Same parabola, three lenses. Use the form that matches what you're being asked.
Which form when?

Need the vertex? → vertex form. Need the roots? → factored form. Need to plug into the quadratic formula? → standard form.

Reading the vertex from vertex form

f(x) = 2(x − 3)2 + 5 Vertex form a(x − h)2 + k → vertex (h, k) Vertex: (3, 5)
Practice

Read the vertex

Find the vertex of f(x) = 2(x − 3)² + 5.

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Completing the square

Goal: convert standard form to vertex form. The trick is to take half of b, square it, and add (and subtract!) it inside.

x2 + 6x + 5 = 0 x2 + 6x = −5    (move c to RHS) x2 + 6x + 9 = −5 + 9    (add (b/2)² = 9 to both sides) (x + 3)2 = 4 x + 3 = ±2 x = −1 or x = −5
Practice

Complete the square

Solve by completing the square: x² + 6x + 5 = 0.

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The discriminant tells you how many roots

Discriminant = b2 − 4ac Positive: two distinct real roots (parabola crosses x-axis twice) Zero: one repeated real root (parabola just touches x-axis) Negative: no real roots — two complex roots (parabola misses x-axis)
Practice

Compute the discriminant

The discriminant of x² + 4x + 5 = 0 is:

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Real-world: projectile motion

A ball is thrown upward with the height equation h(t) = −16t² + 64t + 5 (height in feet, t in seconds). The maximum height is at the vertex.

tmax = −b / (2a) = −64 / (−32) = 2 seconds h(2) = −64 + 128 + 5 = 69 feet
Practice

Maximum height

A ball is thrown so that h(t) = −16t² + 64t + 5. What is the maximum height?

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3-second recap

  • Three forms: standard, vertex, factored. Pick the one that matches the question.
  • Vertex from standard: x = −b / (2a).
  • Completing the square: add (b/2)2 to both sides.
  • Discriminant b2−4ac: + → 2 real, 0 → 1 repeated, − → 2 complex.